a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}\)

\(=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)

\(\sqrt{\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}+3}=\sqrt{\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

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

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

\(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}+\sqrt{5}\)

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

b) Ta có: \(\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{19-8\sqrt{3}+3}}\)

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

=4

a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)

\(=3-2=1\)

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

b: B<A

=>B-1<0

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

=>-1/căn x<0

=>căn x>0

=>x>0 và x<>1

`A=((3sqrtx+6)/(x-4)+sqrtx/(sqrtx-2)):(x-9)/(sqrtx-3)(x>=0,x ne 4,x ne 9)`

`=((3(sqrtx+2))/((sqrtx-2)(sqrtx+2))+sqrtx/(sqrtx-2)):((sqrtx-3)(sqrtx+3))/(sqrtx-3)`

`=(3/(sqrtx-2)+sqrtx/(sqrtx-2)):(sqrtx+3)`

`=(sqrtx+3)/(sqrtx-2)*1/(sqrtx+3)`

`=1/(sqrtx-2)`

\(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

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

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

Ta có: \(A=\left(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{x-9}{\sqrt{x}-3}\)

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

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

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

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

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

b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)

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

\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-6< 0\)

\(\Leftrightarrow x< 36\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)

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 \(A=x\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1; x\neq 25$

a) 

\(A=\frac{4\sqrt{x}}{\sqrt{x}-5}:\left[\frac{(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2}+\frac{5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\right]\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}}{\sqrt{x}-5}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{4(\sqrt{x}+2)}{\sqrt{x}-5}\)

b) Tại $x=81$ thì $\sqrt{x}=9$.

Khi đó: $A=\frac{4(9+2)}{9-5}=11$

c) $A< 4\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}-5}< 1$

$\Leftrightarrow \frac{7}{\sqrt{x}-5}< 0\Leftrightarrow \sqrt{x}-5< 0$

$\Leftrightarrow 0\leq x< 25$. Kết hợp với ĐKXĐ suy ra: $0\leq x< 25; x\neq 1$