\(a)x\ne\pm\frac{4}{3}\)

\(b)x\ne2\)

\(c)x\ne\pm1\)

\(d)x\ne0;x\ne\frac{1}{2}\)

\(e)x\ne\pm1\)

\(f)x\ne-1;x\ne3\)

\(g)x\ne3;x\ne2\)

Mình Không Biết !

Ta có: \(A=\frac{2m^2-4m+5}{m^2-2m+2}\)

\(=\frac{2m^2-4m+2+3}{m^2-2m+1+1}=\frac{2\left(m^2-2m+1\right)+3}{\left(m^2-2m+1\right)+1}\)

\(=\frac{2\left(m-1\right)^2+3}{\left(m-1\right)^2+1}\ge\frac{3}{1}=3\) (do \(\left(m-1\right)^2\ge0\))

Dấu "=" xảy ra \(\Leftrightarrow m-1=0\Leftrightarrow m=1\)

Vậy \(A_{min}=3\Leftrightarrow m=1\)

\(A=2+\frac{1}{m^2-2m+1+1}=2+\frac{1}{\left(m-1\right)^2+1}\)

\(\left(m-1\right)^2+1\ge1\Leftrightarrow\frac{1}{\left(m-1\right)^2+1}\le1\)

\(\Rightarrow A\le3\)

 \("="\Leftrightarrow m=1\)

ĐKXĐ : \(x\ne0;x\ne1\)

\(\frac{a}{x}+\frac{b}{1-x}=\frac{1}{x\left(1-x\right)}\)

\(\frac{a\left(1-x\right)}{x\left(1-x\right)}+\frac{bx}{x\left(1-x\right)}=\frac{1}{x\left(1-x\right)}\)

\(\frac{a\left(1-x\right)+bx}{x\left(1-x\right)}=\frac{1}{x\left(1-x\right)}\)

\(\Rightarrow a\left(1-x\right)+bx=1\)

\(\Leftrightarrow a-ax+bx=1\)

\(\Leftrightarrow a=ax-bx\)

\(\Leftrightarrow a=x\left(a-b\right)\)

\(\Leftrightarrow x=\frac{a}{a-b}\)

a) ĐKXĐ : \(x\ne0;x\ne\pm2;x\ne3\)

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

Đặt \(B=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)

\(B=\frac{\left(x+2\right)\left(x+2\right)}{-\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{\left(2-x\right)\left(x-2\right)}{\left(2+x\right)\left(x-2\right)}\)

\(B=\frac{-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}-\frac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-\left(x+2\right)^2-4x^2--\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(B=\frac{-4x}{x-2}\)

\(\Rightarrow A=\frac{-4x}{x-2}:\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(\Leftrightarrow A=\frac{-4x\cdot x^2\cdot\left(2-x\right)}{\left(x-2\right)\cdot x\cdot\left(x-3\right)}\)

\(\Leftrightarrow A=\frac{4x^2\cdot x\cdot\left(x-2\right)}{\left(x-3\right)\cdot x\cdot\left(x-2\right)}\)

\(\Leftrightarrow A=\frac{4x^2}{x-3}\)

b) \(\left|x-7\right|=4\)

\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}}\)

Mà ĐKXĐ x khác 3 => x = 11

\(\Leftrightarrow A=\frac{4\cdot11^2}{11-3}=\frac{121}{2}\)

c) \(A=\frac{4x^2}{x-3}\)

Để A dương thì hoặc cả tử và mẫu âm hoặc cả tử và mẫu dương

Dễ thấy \(4x^2\ge0\forall x\)

=> Để A dương thì x - 3 dương

hay x - 3 > 0

<=> x > 3

Vậy x > 3 thì A > 0