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

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

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

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

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

\(B=1\)

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

\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)

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

đk : xkhác -1 ; 1 

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

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

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

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

Đề sai rồi bạn

`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`

`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`

`=((x+3\sqrtx-4)/(x-4)).((x-4)/\sqrtx))`

`=(x+3\sqrtx)/\sqrtx`

`=(\sqrtx(\sqrtx+3))/\sqrtx`

`=\sqrtx+3(đpcm)`

`2)P=x+3

`<=>\sqrtx+3=x+3`

`<=>x-\sqrtx=0`

`<=>\sqrtx(\sqrtx-1)=0`

Vì `x>0=>\sqrtx>0`

`=>\sqrtx-1=0<=>x=1(tm)`

Vậy `x=1=>\sqrtx+3=x+3`

`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`

`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`

`=((x+3\sqrtx)/(x-4)).((x-4)/\sqrtx))`

`=(x+3\sqrtx)/\sqrtx`

`=(\sqrtx(\sqrtx+3))/\sqrtx`

`=\sqrtx+3(đpcm)`

`2)P=x+3

`<=>\sqrtx+3=x+3`

`<=>x-\sqrtx=0`

`<=>\sqrtx(\sqrtx-1)=0`

Vì `x>0=>\sqrtx>0`

`=>\sqrtx-1=0<=>x=1(tm)`

Vậy `x=1=>\sqrtx+3=x+3`

1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

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

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

3: 2P=2*căn x+5

=>\(\dfrac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)

=>\(2x+5\sqrt{x}-2\sqrt{x}-2=0\)

=>\(2x+3\sqrt{x}-4=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

=>\(2\sqrt{x}-1=0\)

=>x=1/4

1:

ĐKXĐ: \(x\notin\left\{3;-2;1\right\}\)

 \(A=\left(\dfrac{x\left(x+2\right)-x+1}{\left(x-3\right)\left(x+2\right)}\right):\left(\dfrac{x\left(x-3\right)+5x+1}{\left(x+2\right)\left(x-3\right)}\right)\)

\(=\dfrac{x^2+2x-x+1}{\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(x+2\right)\left(x-3\right)}{x^2-3x+5x+1}\)

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