Wyrażenia wymierne - działania

Zbiór zadań Odsłon: 2645

$$
\text { a) } \begin{aligned}
& \frac{7 a}{x^2-9}+\frac{5 a}{x-3}-\frac{a}{x+3}-2=\frac{7 a}{(x-3)(x+3)}+\frac{5 a}{x-3}-\frac{a}{x+3}-2= \\
= & \frac{7 a+5 a(x+3)-a(x-3)-2(x-3)(x+3)}{(x-3)(x+3)}= \\
= & \frac{7 a+5 a x+15 a-a x+3 a-2\left(x^2-9\right)}{x^2-9}=\frac{25 a+4 a x-2 x^2+18}{x^2-9}
\end{aligned}
$$
b) $$\frac{x^2+y^2}{x^2-y^2}-\frac{x+y}{2 x-2 y}+1=\frac{x^2+y^2}{(x-y)(x+y)}-\frac{x+y}{2(x-y)}+1=$$ $$=\frac{2\left(x^2+y^2\right)-(x+y)^2+2(x-y)(x+y)}{2(x-y)(x+y)}=$$ $$=\frac{2 x^2+2 y^2-\left(x^2+2 x y+y^2\right)+2\left(x^2-y^2\right)}{2\left(x^2-y^2\right)}=$$ $$=\frac{2 x^2+2 y^2-x^2-2 x y-y^2+2 x^2-2 y^2}{2\left(x^2-y^2\right)}=\frac{3 x^2-2 x y-y^2}{2\left(x^2-y^2\right)}$$ Zakładamy, że $$x \neq y$$ i $$x \neq-y$$.

c) $$\frac{7}{2 x-4}-\frac{3}{x+2}-\frac{12}{x^2-4}+x=\frac{7}{2(x-2)}-\frac{3}{x+2}-\frac{12}{(x-2)(x+2)}+x=$$
$$=\frac{7(x+2)-6(x-2)-24+2 x(x-2)(x+2)}{2(x-2)(x+2)}=$$
$$=\frac{7 x+14-6 x+12-24+2 x\left(x^2-4\right)}{2\left(x^2-4\right)}=\frac{x+2+2 x^3-8 x}{2\left(x^2-4\right)}=$$
$$=\frac{2 x^3-7 x+2}{2\left(x^2-4\right)} \quad$$ Zakładamy, że $$x \neq 2$$ i $$x \neq-2$$.
d) $$\frac{7}{8 a^2-18 b^2}+\frac{1}{2 a^2+3 a b}-\frac{1}{4 a b-6 b^2}-2 a=$$
$$=\frac{7}{2\left(4 a^2-9 b^2\right)}+\frac{1}{a(2 a+3 b)}-\frac{1}{2 b(2 a-3 b)}-2 a=$$
$$=\frac{7}{2(2 a-3 b)(2 a+3 b)}+\frac{1}{a(2 a+3 b)}-\frac{1}{2 b(2 a-3 b)}-2 a=$$

$$
\begin{aligned}
& =\frac{7 a b+2 b(2 a-3 b)-a(2 a+3 b)-4 a^2 b(2 a-3 b)(2 a+3 b)}{2 a b(2 a-3 b)(2 a+3 b)}= \\
& =\frac{7 a b+4 a b-6 b^2-2 a^2-3 a b-4 a^2 b\left(4 a^2-9 b^2\right)}{2 a b\left(4 a^2-9 b^2\right)}= \\
& =\frac{8 a b-6 b^2-2 a^2-16 a^4 b+36 a^2 b^3}{2 a b\left(4 a^2-9 b^2\right)}
\end{aligned}
$$
Zakładamy, ze $$a \neq 0, \quad b \neq 0, \quad 2 a \neq 3 b, \quad 2 a \neq-3 b$$.

a)
$$
\begin{aligned}
& \frac{2}{n+2}+\frac{n+3}{n^2-4}-\frac{3 n+1}{n^2-4 n+4}=\frac{2}{n+2}+\frac{n+3}{(n-2)(n+2)}-\frac{3 n+1}{(n-2)^2}= \\
& =\frac{2(n-2)^2+(n+3)(n-2)-(3 n+1)(n+2)}{(n+2)(n-2)^2}= \\
& =\frac{2\left(n^2-4 n+4\right)+n^2-2 n+3 n-6-\left(3 n^2+6 n+n+2\right)}{(n+2)(n-2)^2}= \\
& =\frac{2 n^2-8 n+8+n^2+n-6-3 n^2-6 n-n-2}{(n+2)(n-2)^2}=\frac{-14 n}{(n+2)(n-2)^2}
\end{aligned}
$$
Zakładamy, ze $$n \neq 2$$ i $$n \neq-2$$.
b)
$$
\begin{aligned}
& \frac{3}{2 m+6}-\frac{m-2}{m^2+6 m+9}=\frac{3}{2(m+3)}-\frac{m-2}{(m+3)^2}= \\
& =\frac{3(m+3)-2(m-2)}{2(m+3)^2}=\frac{3 m+9-2 m+4}{2(m+3)^2}=\frac{m+13}{2(m+3)^2}
\end{aligned}
$$
Zakładamy, zee $$m \neq-3$$.
c)
$$
\begin{aligned}
& \frac{1+a}{a-3}-\frac{1-2 a}{3+a}-\frac{a(1-a)}{9-a^2}=\frac{1+a}{a-3}-\frac{1-2 a}{a+3}+\frac{a(1-a)}{a^2-9}= \\
& =\frac{(1+a)(a+3)-(1-2 a)(a-3)+a(1-a)}{a^2-9}= \\
& =\frac{a+3+a^2+3 a-\left(a-3-2 a^2+6 a\right)+a-a^2}{a^2-9}= \\
& =\frac{a^2+4 a+3-a+3+2 a^2-6 a+a-a^2}{a^2-9}=\frac{2 a^2-2 a+6}{a^2-9}= \\
& =\frac{2\left(a^2-a+8\right.}{a^2-9} \text { Zakadamy, ze } a \neq 3 \text { i } a \neq-3
\end{aligned}
$$

$$
\text { d) } \begin{aligned}
& \frac{1}{p-3}-\frac{3}{2 p+6}-\frac{p}{2 p^2-12 p+18}= \\
& =\frac{1}{p-3}-\frac{3}{2(p+3)}-\frac{p}{2\left(p^2-6 p+9\right)}= \\
& =\frac{1}{p-3}-\frac{3}{2(p+3)}-\frac{p}{2(p-3)^2}= \\
& =\frac{2(p-3)(p+3)-3(p-3)^2-p(p+3)}{2(p-3)^2(p+3)}= \\
& =\frac{2\left(p^2-9\right)-3\left(p^2-6 p+9\right)-p^2-3 p}{2(p-3)^2(p+3)}= \\
& =\frac{2 p^2-18-3 p^2+18 p-27-p^2-3 p}{2(p-3)^2(p+3)}=\frac{-2 p^2+15 p-10}{2(p-3)^2(p+3)}
\end{aligned}
$$
Zakładamy, zee $$p \neq 3$$ i $$p \neq-3$$

$$
\text { a) } \begin{aligned}
& \frac{3 x+2}{x^2-2 x+1}-\frac{6}{x^2-1}-\frac{3 x-2}{x^2+2 x+1}= \\
= & \frac{3 x+2}{(x-3)^2}-\frac{6}{(x-1)(x+1)}-\frac{3 x-2}{(x+1)^2}= \\
= & \frac{(3 x+2)(x+1)^2-6(x-1)(x+1)}{(x-1)^2(x+1)^2}-\frac{(3 x-2)(x-1)^2}{(x-1)^2(x+1)^2}= \\
= & \frac{(3 x+2)\left(x^2+2 x+1\right)-6\left(x^2-1\right)}{(x-1)^2(x+1)^2}-\frac{(3 x-2)\left(x^2-2 x+1\right)}{(x-1)^2(x+1)^2}= \\
= & \frac{3 x^3+6 x^2+3 x+2 x^2+4 x+2-6 x^2+6}{(x-1)^2(x+1)^2}+ \\
- & \frac{3 x^3+2 x^2+7 x+8-3 x^3+8 x^2-7 x+2}{(x-1)^2(x+1)^2}=\frac{3 x^3-6 x^2+3 x-2 x^2+4 x-2}{(x-1)^2(x+1)^2}= \\
= & \frac{10 x^2+10}{(x-1)^2(x+1)^2}=
\end{aligned}
$$
Zakladamy, zee $$x \neq 1$$ i $$x \neq-1$$.

$$
\text { b) } \begin{aligned}
& \frac{1}{a-b}-\frac{3 a b}{a^3-b^3}-\frac{b-a}{a^2+a b+b^2}= \\
& =\frac{1}{a-b}-\frac{3 a b}{(a-b)\left(a^2+a b+b^2\right.}-\frac{b-a}{a^2+a b+b^2}= \\
& =\frac{a^2+a b+b^2-3 a b-(b-a)(a-b)}{(a-b)\left(a^2+a b+b^2\right)}= \\
& =\frac{a^2+b^2-2 a b-\left(a b-b^2-a^2+a b\right)}{(a-b)\left(a^2+a b+b^2\right)}= \\
& =\frac{a^2+b^2-2 a b-a b+b^2+a^2-a b}{(a-b)\left(a^2+a b+b^2\right)}=\frac{2 a^2-4 a b+2 b^2}{(a-b)\left(a^2+a b+b^2\right)}= \\
& =\frac{2\left(a^2-2 a b+b^2\right)}{(a-b)\left(a^2+a b+b^2\right)}=\frac{2(a-b)^2}{(a-b)\left(a^2+a b+b^2\right)}=\frac{2(a-b)}{a^2+a b+b^2}
\end{aligned}
$$
$$W$$ tym przypadku wystarczy zalożyć, że $$a \neq b$$, ponieważ:
$$
a^2+a b+b^2=\left(a+\frac{1}{2} b\right)^2+\frac{3}{4} b^2 \geqslant 0
$$
przy czym równość $$\left(a+\frac{1}{2} b\right)^2+\frac{3}{4} b^2=0$$ zachodiz tylko dia $$a=b=0$$.

$$
\text { c) } \begin{aligned}
& \frac{1}{2+4 m+2 m^2}-\frac{1}{1-m^2}+\frac{1}{2-4 m+2 m^2}= \\
& =\frac{1}{2\left(1+2 m+m^2\right)}-\frac{1}{(1-m)(1+m)}+\frac{1}{2\left(1-2 m+m^2\right)}= \\
& =\frac{1}{2(1+m)^2}-\frac{1}{(1-m)(1+m)}+\frac{1}{2(1-m)^2}= \\
& =\frac{(1-m)^2-2(1-m)(1+m)+(1+m)^2}{2(1+m)^2(1-m)^2}= \\
& =\frac{1-2 m+m^2-2\left(1-m^2\right)+1+2 m+m^2}{2(1+m)^2(1-m)^2}= \\
& =\frac{1-2 m+m^2-2+2 m^2+1+2 m+m^2}{2(1+m)^2(1-m)^2}=\frac{4 m^2}{2(1+m)^2(1-m)^2}= \\
& =\frac{2 m^2}{2(1+m)^2(1-m)^2}
\end{aligned}
$$
Zakładamy, ze $$m \neq 1$$ i $$m \neq-1$$

d) $$\begin{aligned} & \frac{2}{3 a+6}-\frac{a-2}{2 a^2+4 a}-\frac{2}{3 a^2+12 a+12}-\frac{4}{3 a(a+2)^2}= \\ = & \frac{2}{3(a+2)}-\frac{a-2}{2 a(a+2)}-\frac{2}{3\left(a^2+4 a+4\right)}-\frac{4}{3 a(a+2)^2}= \\ = & \frac{2}{3(a+2)}-\frac{a-2}{2 a(a+2)}-\frac{2}{3(a+2)^2}-\frac{4}{3 a(a+2)^2}= \\ = & \frac{4 a(a+2)-3(a+2)(a-2)-4 a-8}{6 a(a+2)^2}= \\ = & \frac{4 a^2+8 a-3\left(a^2-4\right)-4 a-8}{6 a(a+2)^2}=\frac{4 a^2+8 a-3 a^2+12-4 a-8}{6 a(a+2)^2}= \\ = & \frac{a^2+4 a+4}{6 a(a+2)^2}=\frac{(a+2)^2}{6 a(a+2)^2}=\frac{1}{6 a}\end{aligned}$$

$$
\text { a) } \begin{aligned}
& \frac{5 x^2-10 x y+5 y^2}{2 x^2-2 x y+2 y^2}: \frac{8 x-8 y}{10 x^3+10 y^3}=\frac{5\left(x^2-2 x y+y^2\right)}{2\left(x^2-x y+y^2\right)} \cdot \frac{10\left(x^3+y^3\right)}{8(x-y)}= \\
& =\frac{5(x-y)^2}{2\left(x^2-x y+y^2\right)} \cdot \frac{5(x+y)\left(x^2-x y+y^2\right)}{4(x-y)}=\frac{25(x-y)(x+y)}{8}= \\
& =\frac{25\left(x^2-y^2\right)}{8}
\end{aligned}
$$
Zakładamy, że $$x \neq y$$ i $$x \neq-y$$.

b) $$\frac{a^2-5 a+6}{a^2+7 a+12} \cdot \frac{a^2+3 a}{a^2-4 a+4}=\frac{(a-2)(a-3)}{(a+3)(a+4)} \cdot \frac{a(a+3)}{(a-2)^2}=\frac{a(a-3)}{(a+4)(a-2)}$$ Zakładamy, że $$a \neq-3, a \neq-4, a \neq 2$$.
c) $$\frac{x^2+2 x-3}{x^2+3 x-10}: \frac{x^2+7 x+12}{x^2-9 x+14}=\frac{(x-1)(x+3)}{(x-2)(x+5)}: \frac{(x+3)(x+4)}{(x-2)(x-7)}=$$ $$=\frac{(x-1)(x+3)}{(x-2)(x+5)} \cdot \frac{(x-2)(x-7)}{(x+3)(x+4)}=\frac{(x-1)(x-7)}{(x+5)(x+4)}$$
Zakładamy, że $$x \neq-4, x \neq-3, x \neq-5, x \neq 2, x \neq 7$$
d) $$\frac{14 x^2+x-3}{14 x^2-13 x+3}: \frac{6 x^2+13 x+5}{10 x^2-11 x+3}=$$ $$=\frac{14\left(x-\frac{3}{7}\right)\left(x+\frac{1}{2}\right)}{14\left(x-\frac{3}{7}\right)\left(x-\frac{1}{2}\right)}: \frac{6\left(x+\frac{1}{2}\right)\left(x+\frac{5}{3}\right)}{10\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)}=$$ $$=\frac{x+\frac{1}{2}}{x-\frac{1}{2}} \cdot \frac{5\left(x-\frac{1}{2}\right)\left(x-\frac{3}{5}\right)}{3\left(x+\frac{1}{2}\right)\left(x+\frac{5}{3}\right)}=\frac{5\left(x-\frac{3}{5}\right)}{3\left(x+\frac{5}{3}\right)}=\frac{5 x-3}{3 x+5}$$
Zakładamy, że $$x \neq-\frac{1}{2}, x \neq-\frac{5}{3}, x \neq \frac{3}{7}, x \neq \frac{1}{2}, x \neq \frac{3}{5}$$.
e) $$\frac{a, b+b c+a c+c^2}{a b-a c-b c+c^2}: \frac{b^2-c^2}{a^2-c^2}=\frac{b(a+c)+c(a+c)}{b(a-c)-c(a-c)}: \frac{b^2-c^2}{a^2-c^2}=$$ $$=\frac{(a+c)(b+c)}{(a-c)(b-c)}: \frac{(b-c)(b+c)}{(a-c)(a+c)}=\frac{(a+c)(b+c)}{(a-c)(b-c)} \cdot \frac{(a-c)(a+c)}{(b-c)(b+c)}=$$ $$=\frac{(a+c)^2}{(b-c)^2}$$
Zakładamy, że $$b \neq c, b \neq-c, a \neq c, a \neq-c$$.

$$
\text { f) } \begin{aligned}
& \frac{a^2-4 a b+4 b^2}{c^2-4 b c+4 b^2}: \frac{a c-2 b c+2 a b-4 b^2}{c^2-4 b^2}= \\
= & \frac{(a-2 b)^2}{(c-2 b)^2}: \frac{c(a-2 b)+2 b(a-2 b)}{(c-2 b)(c+2 b)}= \\
= & \frac{(a-2 b)^2}{(c-2 b)^2} \cdot \frac{(c-2 b)(c+2 b)}{(a-2 b)(c+2 b)}=\frac{a-2 b}{c-2 b}
\end{aligned}
$$
Zakładamy, że $$a \neq 2 b, c \neq-2 b, c \neq 2 b$$

a) $$\left(\frac{5 a}{a+x}+\frac{5 x}{a-x}+\frac{10 a x}{a^2-x^2}\right):\left(\frac{a}{a+x}+\frac{x}{a-x}-\frac{2 a x}{a^2-x^2}\right)=$$
$$=\frac{5 a(a-x)+5 x(a+x)+10 a x}{(a-x)(a+x)} ; \frac{a(a-x)+x(a+x)-2 a x}{(a-x)(a+x)}=$$
$$=\frac{5 a^2-5 a x+5 a x+5 x^2+10 a x}{(a-x)(a+x)}: \frac{a^2-a x+a x+x^2-2 a x}{(a-x)(a+x)}=$$
$$=\frac{5 a^2+10 a x+5 x^2}{(a-x)(a+x)}: \frac{a^2-2 a x+x^2}{(a-x)(a+x)}=$$
$$=\frac{5\left(a^2+2 a x+x^2\right)}{(a-x)(a+x)}: \frac{(a-x)^2}{(a-x)(a+x)}=$$
$$=\frac{5(a+x)^2}{(a-x)(a+x)}: \frac{(a-x)^2}{(a-x)(a+x)}=\frac{5(a+x)}{a-x}: \frac{a-x}{a+x}=$$
$$=\frac{5(a+x)}{a-x} \cdot \frac{a+x}{a-x}=\frac{5(a+x)^2}{(a-x)^2}=5\left(\frac{a+x}{a-x}\right)^2$$

b) $$\begin{aligned} & \left(\frac{b}{a^2+a b}-\frac{2}{a+b}+\frac{a}{b^2+a b}\right):\left(\frac{b}{a}-2+\frac{a}{b}\right)= \\ = & \left(\frac{b}{a(a+b)}-\frac{2}{a+b}+\frac{a}{b(b+a)}\right): \frac{b^2-2 a b+a^2}{a b}= \\ = & \frac{b^2-2 a b+a^2}{a b(a+b)}: \frac{(b-a)^2}{a b}=\frac{(b-a)^2}{a b(a+b)} \cdot \frac{a b}{(b-a)^2}=\frac{1}{a+b}\end{aligned}$$

$$
\text { c) } \begin{aligned}
& \left(1+\frac{a}{x}+\frac{a^2}{x^2}\right) \cdot\left(1-\frac{a}{x}\right) \cdot \frac{x^3}{a^3-x^3}= \\
& =\frac{x^2+a x+a^2}{x^2} \cdot \frac{x-a}{x} \cdot \frac{x^3}{(a-x)\left(a^2+a x+x^2\right)}=\frac{x-a}{a-x}=\frac{-(a-x)}{a-x}=-1
\end{aligned}
$$
Zakładamy, że $$x \neq 0, x \neq a$$.
d)
$$
\begin{aligned}
& \left(\frac{2 a+1}{2 a-1}-\frac{2 a-1}{2 a+1}\right):\left[1:\left(1-\frac{1}{a}+\frac{1}{4 a^2}\right)\right]= \\
& =\frac{(2 a+1)^2-(2 a-1)^2}{(2 a-1)(2 a+1)}:\left(1: \frac{4 a^2-4 a+1}{4 a^2}\right)= \\
& =\frac{4 a^2+4 a+1-\left(4 a^2-4 a+1\right)}{(2 a-1)(2 a+1)}: \frac{4 a^2}{4 a^2-4 a+1}= \\
& =\frac{4 a^2+4 a+1-4 a^2+4 a-1}{(2 a-1)(2 a+1)} \cdot \frac{4 a^2-4 a+1}{4 a^2}= \\
& =\frac{8 a}{(2 a-1)(2 a+1)} \cdot \frac{(2 a-1)^2}{4 a^2}=\frac{2(2 a-1)}{a(2 a+1)}
\end{aligned}
$$
Zakladamy, że $$a \neq 0, a \neq-\frac{1}{2}, a \neq \frac{1}{2}$$.

a) $$\left(\frac{a^2-a b}{a^2 b+b^3}-\frac{2 a^2}{b^3-a b^2+a^2 b-a^3}\right) \cdot\left(1-\frac{b-1}{a}-\frac{b}{a^2}\right)=$$
$$=\left(\frac{a(a-b)}{b\left(a^2+b^2\right)}-\frac{2 a^2}{b^2(b-a)+a^2(b-a)}\right) \cdot \frac{a^2-a(b-1)-b}{a^2}=$$
$$=\left(\frac{a(a-b)}{b\left(a^2+b^2\right)}-\frac{2 a^2}{(b-a)\left(a^2+b^2\right)}\right) \cdot \frac{a^2-a b+a-b}{a^2}=$$
$$=\frac{a(a-b)(b-a)-2 a^2 b}{b(b-a)\left(a^2+b^2\right)} \cdot \frac{a(a-b)+a-b}{a^2}=$$
$$=\frac{a\left(a b-a^2-b^2+a b\right)-2 a^2 b}{b(b-a)\left(a^2+b^2\right)} \cdot \frac{(a-b)(a+1)}{a^2}=$$
$$=\frac{a\left(2 a b-a^2-b^2\right)-2 a^2 b}{b(b-a)\left(a^2+b^2\right)} \cdot \frac{(a-b)(a+1)}{a^2}=$$

$$
\begin{aligned}
& =\frac{2 a^2 b-a\left(a^2+b^2\right)-2 a^2 b}{b(b-a)\left(a^2+b^2\right)} \cdot \frac{(a-b)(a+1)}{a^2}= \\
& =\frac{-a\left(a^2+b^2\right)}{b(b-a)\left(a^2+b^2\right)} \cdot \frac{(a-b)(a+1)}{a^2}=\frac{-a}{b(b-a)} \cdot \frac{(a-b)(a+1)}{a^2}= \\
& =\frac{a}{b(a-b)} \cdot \frac{(a-b)(a+1)}{a^2}=\frac{a+1}{a b}
\end{aligned}
$$
Zakładamy, że $$a \neq 0, b \neq 0, a \neq b$$.
$$
\text { b) } \begin{aligned}
& \left(m+1-\frac{1}{1-m}\right):\left(m-\frac{m^2}{m-1}\right)= \\
= & \frac{(m+1)(1-m)-1}{1-m}: \frac{m(m-1)-m^2}{m-1}= \\
= & \frac{m-m^2+1-m-1}{1-m}: \frac{m^2-m-m^2}{m-1}=\frac{-m^2}{1-m}: \frac{-m}{m-1}= \\
= & \frac{-m^2}{1-m} \cdot \frac{m-1}{-m}=\frac{-m^2}{-(m-1)} \cdot \frac{m-1}{-m}=-m
\end{aligned}
$$
Zakladamy, że $$m \neq 0, m \neq 1$$

$$
\text { c) } \begin{aligned}
& \left(a-\frac{4 a b}{a+b}+b\right):\left(\frac{a}{a+b}-\frac{a}{b-a}-\frac{2 a b}{a^2-b^2}\right)= \\
= & \frac{a(a+b)-4 a b+b(a+b)}{a+b}:\left(\frac{a}{a+b}-\frac{a}{-(a-b)}-\frac{2 a b}{a^2-b^2}\right)= \\
= & \frac{a^2+a b-4 a b+a b+b^2}{a+b}:\left(\frac{a}{a+b}+\frac{a}{a-b}-\frac{2 a b}{a^2-b^2}\right)= \\
= & \frac{a^2-2 a b+b^2}{a+b}: \frac{a(a-b)+a(a+b)-2 a b}{(a-b)(a+b)}= \\
= & \frac{a^2-2 a b+b^2}{a+b}: \frac{a^2-a b+a^2+a b-2 a b}{(a-b)(a+b)}=\frac{(a-b)^2}{a+b}: \frac{2 a^2-2 a b}{(a-b)(a+b)}= \\
= & \frac{(a-b)^2}{a+b}: \frac{2 a(a-b)}{(a-b)(a+b)}=\frac{(a-b)^2}{a+b}: \frac{2 a}{a+b}= \\
= & \frac{(a-b)^2}{a+b} \cdot \frac{a+b}{2 a}=\frac{(a-b)^2}{2 a}
\end{aligned}
$$
Zakładany, że $$a \neq 0, a \neq b, a \neq-b$$.

d) $$\left[\frac{b^2+c^2}{b^2 c^2} \cdot\left(\frac{1}{b^2}-\frac{1}{c^2}\right)-\left(\frac{1}{a^2}-\frac{1}{c^2}\right) \cdot \frac{a^2+c^2}{a^2 c^2}\right]: \frac{a^2 b^2}{a^2 b^2}=$$
$$=\left(\frac{b^2-c^2}{b^2 c^2} \cdot \frac{c^2-b^2}{b^2 c^2}-\frac{c^2-a^2}{a^2 c^2} \cdot \frac{c^2+a^2}{a^2 c^2}\right): 1=\frac{c^4-b^4}{b^4 c^4}-\frac{c^4-a^4}{a^4 c^4}=$$
$$=\frac{a^4\left(c^4-b^4\right)-b^4\left(c^4-a^4\right)}{a^4 b^4 c^4}=\frac{a^4 c^4-a^4 b^4-b^4 c^4+a^4 b^4}{a^4 b^4 c^4}=$$
$$=\frac{a^4 c^4-b^4 c^4}{a^4 b^4 c^4}=\frac{c^4\left(a^4-b^4\right)}{a^4 b^4 c^4}=\frac{a^4-b^4}{a^4 b^4}$$
Zakładamy, że $$a \neq 0, b \neq 0, c \neq 0$$

$$
\begin{aligned}
& \text { a) }\left(\frac{x+1}{1-x}-\frac{1-x}{x+1}-\frac{4 x^2}{x^2-1}\right):\left[-2\left(\frac{1}{x^3+x^2}-\frac{1-x}{x^2}\right)\right]= \\
& =\left(\frac{x+1}{1-x}-\frac{1-x}{x+1}+\frac{4 x^2}{1-x^2}\right):\left[-2\left(\frac{1}{x^2(x+1)}-\frac{1-x}{x^2}\right)\right]= \\
& =\frac{(x+1)^2-(1-x)^2+4 x^2}{(1-x)(1+x)}:\left[-2 \cdot \frac{1-(1+x)(1-x)}{x^2(x+1)}\right]= \\
& =\frac{x^2+2 x+1-\left(1-2 x+x^2\right)+4 x^2}{(1-x)(1+x)}:\left[-2 \cdot \frac{1-\left(1-x^2\right)}{x^2(x+1)}\right]= \\
& =\frac{x^2+2 x+1-1+2 x-x^2+4 x^2}{(1-x)(1+x)}:\left[-2 \cdot \frac{1-1+x^2}{x^2(x+1)}\right]= \\
& =\frac{4 x^2+4 x}{(1-x)(1+x)}: \frac{-2 x^2}{x^2(x+1)}=\frac{4 x(x+1)}{(1-x)(1+x)}: \frac{-2}{x+1}=\frac{4 x}{1-x} \cdot \frac{x+1}{-2}= \\
& =\frac{-2 x(x+1)}{1-x}=2 x \cdot\left(\frac{x+1}{x-1}\right) \\
&
\end{aligned}
$$
Zakładany, że $$x \neq 0, x \neq 1, x \neq-1$$.

b) $$\begin{aligned} & \left(\frac{x}{x y+y^2}+\frac{x^2+y^2}{x^3-x y^2}+\frac{y}{x^2-x y}\right): \frac{x^2-x y+y^2}{x^3+y^3}= \\ = & \left(\frac{x}{y(x+y)}+\frac{x^2+y^2}{x\left(x^2-y^2\right)}+\frac{y}{x(x-y)}\right): \frac{x^2-x y+y^2}{(x+y)\left(x^2-x y+y^2\right)}= \\ = & \frac{x^2(x-y)+y\left(x^2+y^2\right)+y^2(x+y)}{x y\left(x^2-y^2\right)}: \frac{1}{x+y}=\end{aligned}$$

$$
\begin{aligned}
& =\frac{x^3-x^2 y+x^2 y+y^3+y^2 x+y^3}{x y\left(x^2-y^2\right)} \cdot(x+y)= \\
& =\frac{x^3+y^2 x+2 y^3}{x y(x-y)(x+y)} \cdot(x+y)=\frac{x^3+y^2 x+2 y^3}{x y(x-y)}
\end{aligned}
$$
Zakładamy, że $$x \neq 0, y \neq 0, x \neq y, x \neq-y$$.
$$
\text { c) } \begin{aligned}
& \left(\frac{5}{a^2-2 a-a x+2 x}-\frac{1}{8-8 a+2 a^2} \cdot \frac{20-10 a}{x-2}\right): \frac{25}{x^3-8}= \\
= & \left.\frac{5}{a(a-2)-x(a-2)}-\frac{1}{2\left(a^2-4 a+4\right)} \cdot \frac{10(2-a)}{x-2}\right): \frac{25}{x^3-8}= \\
= & \left(\frac{5}{(a-2)(a-x)}-\frac{1}{2(a-2)^2} \cdot \frac{-10(a-2)}{x-2}\right): \frac{25}{x^3-8}= \\
= & \left(\frac{5}{(a-2)(a-x)}+\frac{5}{(a-2)(x-2)}\right) \cdot \frac{x^3-8}{25}= \\
= & \frac{5(x-2)+5(a-x)}{(a-2)(a-x)(x-2)} \cdot \frac{x^3-8}{25}=\frac{5 x-10+5 a-5 x}{(a-2)(a-x)(x-2)} \cdot \frac{x^3-8}{25}= \\
= & \frac{5 a-10}{(a-2)(a-x)(x-2)} \cdot \frac{x^3-8}{25}= \\
= & \frac{5(a-2)}{(a-2)(a-x)(x-2)} \cdot \frac{(x-2)\left(x^2+2 x+4\right)}{25}=\frac{x^2+2 x+4}{5(a-x)}
\end{aligned}
$$
Zakładamy, że $$a \neq 2, x \neq 2, a \neq x$$.

d) $$\begin{aligned} & \left(\frac{3 a}{9-3 x-3 a+a x}-\frac{1}{a^2-9}: \frac{x-a}{3 a^2+9 a}\right) \cdot \frac{x^3-27}{3 a}= \\ = & \left(\frac{3 a}{3(3-x)-a(3-x)}-\frac{1}{0^2-9} \cdot \frac{3 a^2+9 a}{x-a}\right) \cdot \frac{x^3-27}{3 a}= \\ = & \left(\frac{3 a}{(3-x)(3-a)}-\frac{1}{(a-3)(a+3)} \cdot \frac{3 a(a+3)}{x-a}\right) \cdot \frac{x^3-27}{3 a}= \\ = & \left(\frac{3 a}{(3-x)(3-a)}-\frac{3 a}{(a-3)(x-a)}\right) \cdot \frac{x^3-27}{3 a}= \\ = & \left(\frac{3 a}{(3-x)(3-a)}+\frac{3 a}{(3-a)(x-a)}\right) \cdot \frac{x^3-27}{3 a}=\end{aligned}$$

$$
\begin{aligned}
& =\frac{3 a(x-a)+3 a(3-x)}{(3-x)(3-a)(x-a)} \cdot \frac{x^3-27}{3 a}=\frac{3 a x-3 a^2+9 a-3 a x}{(3-x)(3-a)(x-a)} \cdot \frac{x^3-27}{3 a}= \\
& =\frac{-3 a^2+9 a}{(3-x)(3-a)(x-a)} \cdot \frac{x^3-27}{3 a}= \\
& =\frac{3 a(3-a)}{(3-x)(3-a)(x-a)} \cdot \frac{(x-3)\left(x^2+3 x+9\right)}{3 a}=\frac{(x-3)\left(x^2+3 x+9\right)}{(3-x)(x-a)}= \\
& =\frac{(x-3)\left(x^2+3 x+9\right)}{(x-3)(a-x)}=\frac{x^2+3 x+9}{a-x}
\end{aligned}
$$
Zakładamy, że $$a \neq 0, a \neq 3, a \neq-3, x \neq 3, a \neq x$$