a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)

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

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

\(=\dfrac{-4}{\sqrt{x}+2}\)

Lời giải:

a) 

\(A=\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right].\frac{2-\sqrt{x}}{\sqrt{x}}\)

\(=\frac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{-4}{\sqrt{x}+2}\)

b) 

$A< -1\Leftrightarrow \frac{-4}{\sqrt{x}+2}+1< 0$

$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+2}< 0$

$\Leftrightarrow \sqrt{x}-2< 0\Leftrightarrow 0\leq x< 4$

Kết hợp với ĐKXĐ suy ra $0< x< 4$

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

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

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

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

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

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

b/ \(A< -1\)

\(\Leftrightarrow\dfrac{-2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\)

Vậy..

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

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

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

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

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

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

\(a,A=\dfrac{x\left(x+2\right)+\left(2-x\right)\left(x-2\right)+12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+2x+2x-4-x^2+2x+12-10x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-4x+8}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Ta có đẳng thức : (2−3+√3√3+1).(2+3−√3√3−1)=1

xét vế trái ta có :(2−3+√3√3+1).(2+3−√3√3−1)  = 

a) ta co \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)=1\)

b) ta co \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)

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

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

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

Vay \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Ta có : \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

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

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

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

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

B = \(\left[\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

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

`=((x-1))/(x+sqrtx)).((sqrtx-1+2)/(x-1))`

`=(x-1)/(x+sqrtx)*(sqrtx+1)/(x-1)`

`=(x-1)/(sqrtx(sqrtx+1))*1/(sqrtx-1)`

`=1/sqrtx`

1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)

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

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

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

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

2: \(\left(x^2-y^2\right)\cdot C=-8\)

=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)

=>\(\left(x-y\right)^3=-8\)

=>x-y=-2

=>x=y-2

\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)

\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)

\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)

\(=\left(y-1\right)\left(-4y+4\right)+4xy\)

\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)

\(=-4y^2+8y-4+4y^2-8y\)
=-4

1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)

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

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

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

\(=\dfrac{x}{x-5}\)