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

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

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

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

b: Để M>1 thì M-1>0

\(\Leftrightarrow\dfrac{x^2-x+1}{x-1}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

a) ĐKXĐ: x # 0; x # 1; x# -1

M = (x^2)/(x-1)

b) x > 1

a. `M=(x+2)/(x\sqrtx-1)+(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`=(x+2)/( (\sqrtx)^3 -1^3))+(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`= (x+2)/((\sqrtx-1)(x+\sqrtx+1)) + +(\sqrt2+1)/(x+\sqrtx+1)-1/(\sqrtx-1)`

`= ((x+2) +(\sqrt2+1)(\sqrtx-1)-(x+\sqrtx+1))/((\sqrtx-1)(x+\sqrtx+1))`

`=( \sqrt2 (\sqrtx-1))/((\sqrtx-1)(x+\sqrtx+1))`

`= (\sqrt2)/(x+\sqrtx+1)`

b. `x=9 => M=\sqrt2/(9+\sqrt9+1)=\sqrt2/13`

a) Ta có: \(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

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

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

b) Thay x=9 vào M, ta được:

\(M=\dfrac{3}{9+3+1}=\dfrac{3}{13}\)

Rút gọn:

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

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

\(M=\frac{x\left(x+1\right)}{x-1}\cdot\frac{x}{3x^3}\)

\(M=\frac{x+1}{3x\left(x-1\right)}\)

MTC = (x - 2)(x + 2)

\(M=\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x^2+4x}{x^2-4}\)

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

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

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

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

\(=\dfrac{x+2}{x-2}\)

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

b: Để M=7 thì (x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(loại)

Vậy: x=-3

\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Tất cả đều phải tìm điều kiện

a: \(=\dfrac{x+1-4}{x+1}\cdot\dfrac{9-x^2+2x^2+2x-8}{-\left(x-3\right)\left(x+3\right)}\)

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

\(=\dfrac{-x-1}{x+3}\)

b: Khi x=-5 thì \(M=\dfrac{-5-1}{-5+3}=\dfrac{-6}{-2}=3\)

c: Để M nguyên thì -x-1 chia hết cho x+3

=>-x-3+2 chia hết cho x+3

=>\(x+3\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{-2;-4;-5\right\}\)