تأكد من فهمك

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

$\frac{2z^2-4z+2}{z^2-7z+6}$

$\frac{2z-4z+2}{z^2-7z+6}=\frac{(2z-2)(z-1)}{(z-6)(z-1)}=\frac{2z-2}{z-6}$

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

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

$\frac{5x+3}{x+3}\times \frac{x^2+5x+6}{25x^2-9}$

$\begin{array}{rl}\frac{5x+3}{x+3}\times \frac{(x^2+5x+6)}{25x^2-9}& =\frac{(5x+3)}{(x+3)}\times \frac{(x+2)(x+3)}{(5x-3)(5x+3)}\\ & =\frac{x+2}{5x-3}\end{array}$

$\frac{z^2+7z-8}{z-1}\times \frac{z^2-4}{z^2+6z-16}$

$\begin{array}{rl}\frac{z^2+7z-8}{z-1}\times \frac{z^2-4}{z^2+6z-16}& =\frac{(z+8)(z-1)}{z-1}\times \frac{(z-2)(z+2)}{(z+8)(z-2)}\\ & =z+2\end{array}$

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

$\begin{array}{rl}\frac{x^2-9}{x^2-4x+4}\times \frac{x^2-4}{x^2-x-6}& =\frac{(x-3)(x+3)}{(x-2{)}^2}\times \frac{(x-2)(x+2)}{(x-3)(x+2)}\\ & =\frac{x+3}{x-2}\end{array}$

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

$\begin{array}{rl}\frac{2y^2-2y}{y^2-9}÷\frac{y^2+y-2}{y^2+2y-3}& =\frac{2y(y-1)}{(y-3)(y+3)}\times \frac{(y+3)(y-1)}{(y+2)(y-1)}\\ & =\frac{2y(y-1)}{(y-3)(y+2)}\end{array}$

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

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

$\begin{array}{rl}\frac{2}{x^2-9}+\frac{3}{x^2-4x+3}& =\frac{2}{(y-3)(x+3)}+\frac{3}{(x-3)(x-1)}\\ & =\frac{2(x-1)+3(x+3)}{(x-3)(x+3)(x-1)}=\frac{2x-2+3x+9}{(x-3)(x+3)(x-1)}=\frac{5x+7}{(x-3)(x+3)(x-1)}\end{array}$

$\frac{2y^3-128}{y^3+4y^2+16y}-\frac{y-1}{y}$

$\begin{array}{rl}\frac{2y^3-128}{y^3+4y^2+16y}-\frac{y-1}{y}& =\frac{2(y-4)(y^2+4y+16)}{y(y^2+4y+16)}-\frac{y-1}{y}\\ & =\frac{2y-8}{y}-\frac{y-1}{y}=\frac{2y-8-y+1}{y}=\frac{y-7}{y}\end{array}$

$\frac{z^2+z+1}{z^4-z}-\frac{z+3}{z^2+2z-3}$

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

$\frac{x^2-1}{x^2-2x+1}-1$

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

$\frac{3}{z-1}+\frac{2}{z+3}+\frac{8}{z^2+2z-3}$

$\begin{array}{c}\frac{3}{z-1}+\frac{2}{z+3}+\frac{8}{z^2+2z-3}=\frac{3}{z-1}+\frac{2}{z+3}+\frac{8}{(z-1)(z+3)}\\ =\frac{3(z+3)+2(z-1)+8}{(z-1)(z+3)}=\frac{3z+9+2z-2+8}{(z-1)(z+3)}=\frac{5z+15}{(z-1)(z+3)}\\ =\frac{5(z+3)}{(z-1)(z+3)}=\frac{5}{(z-1)}\end{array}$

$\frac{y-3}{y-1}+\frac{5y-15}{(y-3{)}^2}-\frac{3y+1}{y^2-4y+3}$

$\begin{array}{rl}\frac{y-3}{y-1}+\frac{5y-15}{(y-3{)}^2}-\frac{3y+1}{(y^2-4y+3)}& =\frac{y-3}{y-1}+\frac{5(y-3)}{(y-3{)}^2}-\frac{3y+1}{(y-3)(y-1)}\\ & =\frac{(y-3{)}^2+5(y-1)-3y-1}{(y-1)(y-3)}\\ & =\frac{y^2-6y+9+5y-5-3y-1}{(y-1)(y-3)}=\frac{y^2-4y+3}{(y-1)(y-3)}=1\end{array}$