\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)

\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{x-1}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{x-1}{2x+1}\\ =\dfrac{x+2}{\left(x+1\right)\left(2x+1\right)}\)

Đề sai r bn

VT = `[ 2/(3x) -2/(x+1) (x+1)/(3x) -x-1)]: (x-1)/x`

`=[2/(3x)-2/(x+1) . ((x+1)-3x(x+1))/(3x) ] . x/(x-1)`

`= [2/(3x)  + 2/(x+1) ((3x-1)(x+1))/(3x) ] . x/(x-1)`

`= [ 2/(3x) + (2(3x-1))/(3x) ] . x/(x-1)`

`= (6x)/(3x) . x/(x-1)`

`= 2 . x/(x-1)`

`= (2x)/(x-1)`

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

Do \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1>0;\forall x\in D\)

\(\Rightarrow\dfrac{-3}{\sqrt{x}+1}< 0\)

\(\Rightarrow A-1< 0\Rightarrow A< 1\)

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

Ta có: \(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2:\dfrac{x^2+y^2}{\left(x-y\right)^2}\)

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

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

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

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

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

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

\(=\dfrac{2xy}{x^2+y^2}\)

Có thể đề bắt cm : VT \(\le1\)

áp dụng kq của bạn thịnh : VT = \(\dfrac{2xy}{x^2+y^2}\le\dfrac{2xy}{2xy}=1\)   (x² + y² \(\ge\) 2xy)

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

a: ĐKXĐ: x<>0; x<>1

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

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

b: |2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=-1(nhận) hoặc x=2(nhận)

Khi x=-1 thì \(P=\dfrac{-1+1}{\left(-1\right)^2}=0\)

Khi x=2 thì \(P=\dfrac{2+1}{2^2}=\dfrac{3}{4}\)

\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\left(dkxd:x\ne1,x\ge0\right)\)

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

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

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

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

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

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

\(B=\dfrac{2x+2\sqrt{x}-3\sqrt{x}-3+3-\sqrt{x}}{x-1}=\dfrac{2x-2\sqrt{x}}{x-1}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\) ĐK: \(x\ge0;x\ne1\)

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

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

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

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

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\) (Đpcm).