Cảm ơn bạn nhé 

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

a) \(A=2x^2-\dfrac{1}{3}y\)

A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)

A=\(\dfrac{5}{3}\)\(x^2y\)

Tại \(x=2;y=9\) ta có

A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60

Vậy tại \(x=2;y=9\) biểu thức A= 60

b) P=\(2x^2+3xy+y^2\)            (\(y^2\) là 1\(y^2\) nha bạn)

P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)

P= 6\(x^3y^3\)

Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có

P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)

Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)

c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)

=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)

=\(-\dfrac{1}{3}\)\(x^4y^2\)

Tại \(x=2;y=\dfrac{1}{4}\)ta có

\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)

\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)

CHÚC BẠN HỌC TỐT NHA

\(ĐK:x\ne-1\\ \left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Với \(x=2\Leftrightarrow A=\dfrac{3}{2+1}=1\)

Với \(x=-2\Leftrightarrow A=\dfrac{3}{-2+1}=-3\)

\(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{3}{2+1}=\dfrac{3}{3}=1\\A=\dfrac{3}{-2+1}=\dfrac{3}{-1}=-3\end{matrix}\right.\)

mình cần gấp mong các bạn giải giùm

c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)

\(=\left[\left(\dfrac{-\left(x-y\right)}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}\right):\dfrac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}\right]:\dfrac{x+1}{2x^2+y+2}\)

\(=\dfrac{-x^2+y^2-x^2-y^2-y+2}{\left(x-2y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)

\(=\dfrac{-2x^2-y+2}{\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)}{\left(2x^2+y-2\right)\left(2x^2+y+2\right)}\cdot\dfrac{2x^2+y+2}{x+1}\)

\(=\dfrac{-1}{x-2y}\)

Thay $x=-1,76$ và $y=\dfrac{3}{25}$ vào $P=\dfrac{-1}{x-2y}$, ta được:

$P=\dfrac{-1}{-1,76-2.(\dfrac{3}{25})}=\dfrac{1}{2}$.