bạn bị tố cáo nha 

đk  x khác 9, x >= 0

\(p=\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{5\sqrt{x}-3}{x-9}\)

\(p=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}-\frac{5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(p=\frac{x+2\sqrt{x}-3-5\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(p=\frac{\sqrt{x}}{\sqrt{x}+3}\)

b, P.(căn x + 3) = |x - 2|

có P = căn x/ căn x + 3

=> căn x = |x - 2|

=> x = |x - 2|^2

=> x = x^2 - 4x + 4

=> x^2 - 5x + 4 = 0

=> (x-1)(x-4) = 0

=> x = 1 hoặc x = 4 (tm)

vậy x = 1 hoặc x = 4

a, \(A=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}\right):\frac{-\sqrt{x}}{x-2\sqrt{x}}\)

\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)

=> 2cawn x + 4 = 12

=> 2.căn x = 8

=> căn x = 4

=> x = 16 (thỏa mãn)

c, có A = 4/ căn x + 2 và B  = 1/căn x - 2

=> A.B = 4/x - 4 

mà AB nguyên

=> 4 ⋮ x - 4

=> x - 4 thuộc Ư(4) 

=> x - 4 thuộc {-1;1;-2;2;-4;4}

=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4

=> x thuộc {3;5;2;6;8}

d, giống c thôi

\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)

\(=\sqrt{72.2}-3\sqrt{24.2}-5\sqrt{8.2}+4.\sqrt{27}\)

\(=12-3\sqrt{3.4^2}-5.4+4.\sqrt{3.3^2}\)

\(=-8-3.4.\sqrt{3}+4.3.\sqrt{3}=-8\)

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

\(Q=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(Q=\frac{x-3\sqrt{x}-2-5\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(Q=\frac{x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)

ủa sao không thấy gọn ta

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

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

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

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

Đặt \(\hept{\begin{cases}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=b\end{cases}\Rightarrow\hept{\begin{cases}a^2+ab+b^2=1\\a-b=2\end{cases}}}\)\(\Rightarrow a=b+2\text{ thế vào phương trình đầu ta có :}\)

\(\left(b+2\right)^2+b\left(b+2\right)+b^2=1\Leftrightarrow3b^2+6b+3=0\Leftrightarrow b=-1\Leftrightarrow\sqrt[3]{3x-1}==-1\)

\(\Leftrightarrow3x-1=-1\Leftrightarrow x=0\)

Minh Quang ơi tại sao \(3x+1-\left(3x-1\right)=\sqrt[3]{3x+1}-\sqrt[3]{3x-1}\)

\(\Leftrightarrow\sqrt[3]{3x+1}-\sqrt[3]{3x-1}=1\)

\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)ĐK : x >= 2 

\(\Leftrightarrow\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=1\)

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

\(\Leftrightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\)

\(\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\Leftrightarrow x=11\)(tmđk)

ĐK : x >= 2

\(\Leftrightarrow\sqrt{\left(x-2\right)-4\sqrt{x-2}+4}+\sqrt{\left(x-2\right)-6\sqrt{x-2}+9}=1\)

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

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

Với 2 ≤ x < 4 (1) trở thành \(2-\sqrt{x-2}+3-\sqrt{x-2}=1\Leftrightarrow-2\sqrt{x-2}=-4\Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x=6\left(ktm\right)\)

Với 4 ≤ x < 11 (1) trở thành \(\sqrt{x-2}-2+3-\sqrt{x-2}=1\Leftrightarrow1=1\left(luondung\right)\)(2)

Với x ≥ 11 (1) trở thành \(\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x=11\)( tm ) (3)

Từ (2) và (3) => S = { x | 4 ≤ x ≤ 11 }