B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)

\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)

b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)

\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))

\(\Leftrightarrow x>-1\).

-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị

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

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b: \(D=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)

Dấu '=' xảy ra khi x=1/4

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

\(A=\left(\dfrac{1}{x+\dfrac{4}{x}+\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}\right)^2\le\left(\dfrac{1}{x+\dfrac{4}{x}+\dfrac{11}{4}}\right)^2\)

\(A\le\left(\dfrac{1}{2\sqrt{\dfrac{4x}{x}}+\dfrac{11}{4}}\right)^2=\dfrac{16}{729}\)

Dấu "=" xảy ra khi \(x=2\)

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

a) Ta có: \(A=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\cdot\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

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

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

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

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

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

b) Thay \(x=-\dfrac{1}{2}\) vào biểu thức \(A=\dfrac{x}{x^2+1}\), ta được:

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

\(\Leftrightarrow A=-\dfrac{1}{2}:\left(\dfrac{1}{4}+1\right)\)

\(\Leftrightarrow A=-\dfrac{1}{2}:\dfrac{5}{4}\)

\(\Leftrightarrow A=-\dfrac{1}{2}\cdot\dfrac{4}{5}\)

\(\Leftrightarrow A=\dfrac{-4}{10}\)

hay \(A=\dfrac{-2}{5}\)

Vậy: Khi \(x=-\dfrac{1}{2}\) thì \(A=\dfrac{-2}{5}\)

c) Để 2A=1 thì \(A=\dfrac{1}{2}\)

hay \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)

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

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

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(không nhận)

Vậy: Không có giá trị nào của x để 2A=1