a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

b: \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)

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

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

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

c: \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=x+1+\dfrac{1}{x-1}\)

=>\(A=x-1+\dfrac{1}{x-1}+2>=2\cdot\sqrt{\left(x-1\right)\cdot\dfrac{1}{x-1}}+2=2+2=4\)

Dấu '=' xảy ra khi (x-1)²=1

=>x-1=1 hoặc x-1=-1

=>x=0(loại) hoặc x=2(nhận)

Vậy: \(A_{min}=4\) khi x=2

a) đặt mẫu chứng là x-2

a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)

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

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

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow2x^2-x+1=0\)

hay \(x\in\varnothing\)

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)

\(=\dfrac{-x^2-4x-4-2x^2+4x-4x^2}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}\)

\(=\dfrac{-7x^2-4}{\left(x+2\right)}\cdot\dfrac{-1}{x-3}=\dfrac{7x^2+4}{\left(x+2\right)\left(x-3\right)}\)

b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)

Lời giải:
ĐK: $x\neq \pm 2; x\neq 0$

a) 

\(A=\left[\frac{x+2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}\right].\frac{2-x}{x}=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}\)

\(=\frac{4x}{(x-2)(x+2)}.\frac{-(x-2)}{x}=\frac{-4}{x+2}\)

b) Để $A=1\Leftrightarrow \frac{-4}{x+2}=1$

$\Leftrightarrow x+2=-4$

$\Leftrightarrow x=-6$ (thỏa ĐKXĐ)

Vậy $x=-6$