ĐK : \(x\ge2,y\ge3,z\ge4\) .

\(pt\Leftrightarrow x+y+z-6=2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-4}\)

\(\Leftrightarrow\left[\left(x-2\right)-2\sqrt{x-2}+1\right]+\left[\left(y-3\right)-2\sqrt{y-3}+1\right]+\left[\left(z-4\right)-2\sqrt{z-4}+1\right]=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-4}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\left(TM\right)\)

a ) ĐK : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)\(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^{^2}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

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

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

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

\(\dfrac{2\sqrt{X}-9}{x-5\sqrt{X}+6}-\dfrac{\sqrt{X}+3}{\sqrt{X}-2}-\dfrac{2\sqrt{X}+1}{3-\sqrt{X}}\) \(\left(X\ne2;X\ne3,X\ge0\right)\)

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

\(=\dfrac{2\sqrt{X}-9-X+9+2X-4\sqrt{X}+\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)

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

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

\(C=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}< 1\)

\(\Rightarrow\dfrac{\sqrt{X}+1-\sqrt{X}+3}{\sqrt{X}-3}< 0\)

\(\Rightarrow\dfrac{4}{\sqrt{X}+3}< 0\) ( VÔ LÍ)

⇒ Không có X thỏa mãn

\(B=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

B=2/3A

=>3căn x/căn x+2=2/3*3=2

=>3căn x=2căn x+4

=>x=16

Bài 2: 

a) Thay m=3 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}x-2y=7\\2x+y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=14\\2x+y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-5y=5\\x-2y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=7+2y=5\end{matrix}\right.\)

Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là (x,y)=(5;-1)

a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)

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

\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)

b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)

\(\Leftrightarrow x+2\sqrt{x}=0\)

hay x=0

3:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

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

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

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

b: M>1/3

=>M-1/3>0

=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)

=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)

=>\(3-\sqrt{x}>0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ

=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ

Dấu = xảy ra khi x=0

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

ĐKXĐ: \(x\ge0;x\ne1\)

Sửa lại đề chỗ \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}\) thành \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

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

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

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

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

\(P=\dfrac{2\sqrt{x}+4-\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)

Để P lớn nhất \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\) lớn nhất

Mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)

\(\Rightarrow P_{max}=1+\dfrac{3}{2}=\dfrac{5}{2}\) khi \(\sqrt{x}+2=2\Leftrightarrow x=0\)