1.

\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)

\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)

2.

\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)

\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)

3.

\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)

\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\\ =\dfrac{x+2-\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}+1-x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)

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

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

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

\(A=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)

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

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

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

\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}+2}\)

\(=\dfrac{5\left(4\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)

ĐK: `x>0` 

`A=((\sqrtx)/(\sqrtx+2) - 4/(x+2\sqrtx)):(1+1/(\sqrtx))`

`=((\sqrtx)/(\sqrtx+2)-4/(\sqrtx(\sqrtx+2))):((\sqrtx+1)/(\sqrtx))`

`=(x -4)/(\sqrtx(\sqrtx+2)) . (\sqrtx)/(\sqrtx+1)`

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

`=(\sqrt-2)/(\sqrtx+1)`

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

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

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

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

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

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

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

a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

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

\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)

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

b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)  (*)

Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)

Bạn santa làm sai r nha!

a, ĐKXĐ: x \(\ge\) 0; x \(\ne\) 4; x \(\ne\) 0

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

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

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

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

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

B = \(\dfrac{2-\sqrt{x}}{3\sqrt{x}}\) (Đoạn này bạn kia viết sai đề mà vẫn đúng kết quả được?)

Vậy ...

b, Ta có: x = 4 + 2\(\sqrt{3}\) = (\(\sqrt{3}\) + 1)² (TMĐK)

\(\Rightarrow\) \(\sqrt{x}\) = \(\sqrt{3}+1\) (1)

Thay (1) vào B ta được:

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

Vậy ...

Chúc bn học tốt!

mình làm lại nhé :

đkxđ : \(\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)

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

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

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

câu b làm như kia là oke rồi nhé <3