تدرب وحل التمرينات

(1)- اكتب كل مقدار من المقادير التالية بأبسط صورة:

$\frac{x+5}{12x}\times \frac{6x-30}{x^2-25}$

$\frac{x+5}{12x}\times \frac{6x-30}{x^2-25}=\frac{x+5}{12x}\times \frac{6(x-5)}{(x-5)(x+5)}=\frac{1}{2x}$

$\frac{y+3}{2y^2+6y+18}\times \frac{y^3-27}{y^2-9}$

$\frac{y+3}{(2y^2+6y+18)}\times \frac{y^3-27}{y^2-9}=\frac{y+3}{2(y^2+3y+9)}\times \frac{(y-3)(y^2+3y+9)}{(y-3)(y+3)}=\frac{1}{2}$

$\frac{3-x}{4-2x}\times \frac{x^2+x-6}{9-x^2}$

$\frac{3-x}{4-2x}\times \frac{x^2+x-6}{9-x^2}=\frac{3-x}{-2(x-2)}\times \frac{(x+3)(x-2)}{(3-x)(3+x)}=\frac{-1}{2}$

$\frac{y+2}{2y-4}÷\frac{y^3+8}{y-2}$

$\begin{array}{rl}\frac{y+2}{2y-4}÷\frac{y^3+8}{y-2}& =\frac{(y+2)}{2(y-2)}\times \frac{y-2}{(y+2)(y^2-2y+4)}\\ & =\frac{1}{2(y^2-2y+4)}\end{array}$

$\frac{y^2-7y}{y^3-27}÷\frac{y^2-49}{y^2+3y+9}$

$\begin{array}{rl}\frac{y^2-7y}{y^3-27}÷\frac{y^2-49}{y^2+3y+9}& =\frac{y(y-7)}{(y-3)(y^2+3y+9)}\times \frac{y^2+3y+9}{(y-7)(y+7)}\\ & =\frac{y}{(y-3)(y+7)}\end{array}$

$\frac{64-z^3}{32+8z+2z^2}÷\frac{(4-z{)}^2}{16-z^2}$

$\frac{64-z^3}{32+8z+2z^2}÷\frac{(4-z{)}^2}{16-z^2}=\frac{(4-z)(16+4z+z^2)}{2(16+4z+z^2)}\times \frac{(4-z)(4+z)}{(4-z{)}^2}=\frac{4+z}{2}$

(2)- اكتب كل مقدار من المقادير التالية بأبسط صورة:

$\frac{5}{x^2-36}-\frac{2}{x^2-12x+36}$

$\begin{array}{rl}& \frac{5}{x^2-36}-\frac{2}{x^2-12x+36}=\frac{5}{(x-6)(x+6)}-\frac{2}{(x-6{)}^2}\\ & =\frac{5(x-6)-2(x+6)}{(x-6{)}^2(x+6)}=\frac{5x-30-2x-12}{(x-6{)}^2(x+6)}=\frac{3x-42}{(x-6{)}^2(x+6)}\end{array}$

$\frac{y^2-y}{y^3-1}-\frac{1}{y^2+y+1}$

$\frac{y^2-y}{y^3-1}-\frac{1}{y^2+y+1}=\frac{y(y-1)}{(y-1)(y^2+y+1)}-\frac{1}{(y^2+y+1)}=\frac{y-1}{(y^2+y+1)}$

$\frac{3}{x-2}-\frac{2}{x-2}+\frac{4+2x+x^2}{x^3-8}$

$\begin{array}{rl}\frac{3}{x-2}-\frac{2}{x-2}+\frac{4+2x+x^2}{x^3-8}& =\frac{1}{x-2}+\frac{4+2x+x^2}{(x-2)(x^2+2x+4)}\\ & =\frac{1+1}{x-2}=\frac{2}{x-2}\end{array}$

$\frac{y-5}{y+1}+\frac{y-1}{y+5}-\frac{25}{y^2+6y+5}$

$\begin{array}{rl}& \frac{y-5}{y+1}+\frac{y-1}{y+5}-\frac{25}{y^2+6y+5}=\frac{y-5}{y+1}+\frac{y-1}{y+5}-\frac{25}{(y+1)(y+5)}\\ & =\frac{(y-5)(y+5)+(y-1)(y+1)-25}{(y+1)(y+5)}=\frac{y^2-25+y^2-1-25}{(y+1)(y+5)}=\frac{2y^2-51}{(y+1)(y+5)}\end{array}$