a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì

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

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

c: M>0

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\)

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4

a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)

c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

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

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

b) \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{4}+1}{\sqrt{4}-1}=\dfrac{2+1}{2-1}=3\)

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

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

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

Bài 1 : 

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

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)

a) Ta có: \(M=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

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

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

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

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

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

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

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

b: M=20

=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)

=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)

=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)

=>\(x\in\varnothing\)

\(M=\dfrac{x^3\left(x^2+x\sqrt[3]{6}+\sqrt[3]{36}\right)}{\left|x^3-3\right|-3}=\dfrac{48\left(4\sqrt[3]{36}+2\sqrt[3]{36}+\sqrt[3]{36}\right)}{48-3-3}\\ M=\dfrac{48\cdot7\sqrt[3]{36}}{42}=8\sqrt[3]{36}\)

mik cần gấp ạ^^

a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

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

b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)

c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9

`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

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

\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

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

`b,`Ta có :

 \(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)

\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)

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

\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)

\(M=\dfrac{6+9\sqrt{2}}{2}\)

`c,`  Để `M<1` Ta có :

 \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)

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

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

\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )

\(\sqrt{x}< 3\)

\(x< 9\)

Đối chiếu ĐKXĐ ta có : `0<x<9`