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

\(=3x+6\sqrt{x}-\left(x-1\right)\)

\(=3x+6\sqrt{x}-x+1\)

\(=2x+6\sqrt{x}+1\)

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

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

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

\(=-x+8\sqrt{x}+1\)

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

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

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

\(=4x-3\sqrt{x}-3\)

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

\(=x-9-\left(2x-3\sqrt{x}-2\right)\)

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

\(=-x+3\sqrt{x}-7\)

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

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

\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)

\(=x-4-4x-6\sqrt{x}+4\)

\(=-3-6\sqrt{x}\)

`(\sqrt(3x^2-12x+12)-x+2)/(x-2)`

`=(\sqrt(3(x^2-4x+4))-(x-2))/(x-2)`

`=(\sqrt(3(x-2)^2)) -(x-2))/(x-2)`

`=(\sqrt3. (x-2) - (x-2))/(x-2)`

`=( (\sqrt3-1) (x-2))/(x-2)`

`=\sqrt3-1`

`=>` C.

\(x\ge0,x\ne9\)

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

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

\(A=\left[\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right].\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(A=\dfrac{-3\left(\sqrt{x}+1\right).\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)

mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)

ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)

  <=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)

<=> \(A^3=x^3-3x+3A\)

<=> \(A^3-3A-x^3+3x=0\)

<=>\(\left(A^3-x^3\right)-3A+3x=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)

<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )

vậy \(A=x\)

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

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

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

b, với x > 0 

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

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

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: $a\geq 0; a\neq 1$

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

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

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

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

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

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

Vậy \(B=\dfrac{2}{\sqrt{x}+3}\)

a

ĐK: \(1< x\ne10\)

Đặt \(t=\sqrt{x-1}\Rightarrow x=t^2+1;0< t\ne3\)

Khi đó:

\(P=\left(\dfrac{t}{3+t}+\dfrac{t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t^2-3t}-\dfrac{1}{t}\right)\\ =\left(\dfrac{t\left(3-t\right)+t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t\left(t-3\right)}-\dfrac{1}{t}\right)\\ =\dfrac{3t+9}{\left(3-t\right)\left(3+t\right)}:\dfrac{3t+1-t+3}{t\left(t-3\right)}=\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}:\dfrac{2t+4}{t\left(t-3\right)}\\ =\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}.\dfrac{t\left(t-3\right)}{2t+4}=\dfrac{-3t}{2t+4}=\dfrac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)

b

Ta có:

\(x=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{5}\left|1-\sqrt{2}\right|\)

\(=\sqrt{2}+1-\left(\sqrt{5}+1\right)\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\)

\(=\sqrt{2}+1-\sqrt{5}\left|1-\sqrt{2}\right|-\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\\ =\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)

Vậy \(P=\dfrac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\dfrac{1}{2}\)