\(y^2-\left(y+2\right)x^2=1\Rightarrow y^2-yx^2-2x^2=1\)\(\Rightarrow y^2\left(1-x^2\right)-2\left(x^2-1\right)=3\)

\(\Rightarrow\left(x^2-1\right)\left(y^2+2\right)=-3\).Mà \(x^2-1>-2,y^2+2>1\)

nên ta có bảng

Vậy \(\left(x,y\right)\in[\left(0,1\right),\left(0,-1\right)]\)

a) Ta có: \(x=9\)thỏa mãn đk 

\(\Rightarrow\)Thay \(x=9\)vào biểu thức ta được: 

\(A=\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=\frac{-9}{2}\)

b) Với x thỏa mãn ĐKXĐ thì ta có:

\(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}\)

\(=\frac{1}{\sqrt{x}+2}+\frac{x+14}{x-4}-\frac{4}{\sqrt{x}-2}\)

\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-2\right)+\left(x+12\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-2+x+12-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

1. x = 9 => A = \(\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=-\frac{9}{2}\)

2. \(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}=\frac{\sqrt{x}-2+x+12-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

3. \(AB>-\frac{3}{4}\) <=> \(\frac{3\sqrt{x}}{1-\sqrt{x}}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+2}>-\frac{3}{4}\)

<=> \(-\frac{3\sqrt{x}}{\sqrt{x}+2}+\frac{3}{4}>0\)

<=> \(\frac{12\sqrt{x}-3\sqrt{x}-4}{4\left(\sqrt{x}+2\right)}< 0\)

<=> \(\frac{9\sqrt{x}-4}{4\sqrt{x}+8}< 0\)

Do \(4\sqrt{x}+8>0\)với mọi x => \(9\sqrt{x}-4< 0\) <=> \(x< \frac{16}{81}\)

Điều kiện để A xác định: \(x\ge0\)

\(A=\frac{2\sqrt{x}+7}{\sqrt{x}+2}=\frac{2\sqrt{x}+4+3}{\sqrt{x}+2}=\frac{2\left(\sqrt{x}+2\right)+3}{\sqrt{x}+2}=2+\frac{3}{\sqrt{x}+2}\)

Vì \(\sqrt{x}+2\)luôn xác định \(\Rightarrow\sqrt{x}+2\ge2\)

\(\Rightarrow\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\)\(\Rightarrow2+\frac{3}{\sqrt{x}+2}\le2+\frac{3}{2}=\frac{7}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x=0\)

Vậy \(maxA=\frac{7}{2}\Leftrightarrow x=0\)

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

Để A đạt GTLN \(\Leftrightarrow\frac{3}{\sqrt{x}+2}\)đạt GTLN

\(\frac{3}{\sqrt{x}+2}\)có mẫu dương, tử dương và tử không đổi nên \(\frac{3}{\sqrt{x}+2}\)đạt GTLN \(\Leftrightarrow\sqrt{x}+2\)đạt GTNN

Có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\).Dấu bằng đẳng thức xảy ra khi \(x=0\).

Khi đó: \(\frac{3}{\sqrt{x}+2}=\frac{3}{2}\)và \(A=2+\frac{3}{2}=\frac{7}{2}\)

               KL: \(A_{max}=\frac{7}{2}\Leftrightarrow x=0\)

\(M=\frac{3\sqrt{x}-2}{\sqrt{x}+2}=\frac{3\left(\sqrt{x}+2\right)-8}{\sqrt{x}+2}=1-\frac{8}{\sqrt{x}+2}\ge1\)

Min M = 1 \(\Leftrightarrow\frac{8}{\sqrt{x}+2}=0\left(1\right)\)

Mà \(\hept{\begin{cases}8>0\\\sqrt{x}+2>0\forall x\end{cases}\Leftrightarrow\frac{8}{\sqrt{x}+2}>0\left(2\right)}\)

Từ ( 1 ) và ( 2 ) => Không tìm được GTNN của M thỏa mãn  \(M=\frac{3\sqrt{x}-2}{\sqrt{x}+2}\)

               Bài làm :

\(a\text{)}\sqrt{x-3}=5\Leftrightarrow\sqrt{x-3}=\sqrt{25}\Leftrightarrow x-3=25\Leftrightarrow x=28\)

\(b\text{)}\sqrt{3x}=3\sqrt{5}\Leftrightarrow\sqrt{3x}=\sqrt{9}.\sqrt{5}\Leftrightarrow\sqrt{3x}=\sqrt{45}\Leftrightarrow3x=45\Leftrightarrow x=\frac{45}{3}=15\)

\(c\text{)}\sqrt{x^2+2x+1}=10\Leftrightarrow\sqrt{\left(x+1\right)^2}=\sqrt{100}\Leftrightarrow\left(x+1\right)^2=10\Leftrightarrow\orbr{\begin{cases}x+1=10\\x+1=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-11\end{cases}}\)

a) \(ĐKXĐ:x\ge3\)

\(\sqrt{x-3}=5\)\(\Leftrightarrow\left(\sqrt{x-3}\right)^2=5^2\)

\(\Leftrightarrow x-3=25\) \(\Leftrightarrow x=28\)( thỏa mãn )

Vậy \(x=28\)

b) \(ĐKXĐ:x\ge0\)

\(\sqrt{3x}=3\sqrt{5}\)\(\Leftrightarrow\left(\sqrt{3x}\right)^2=\left(3\sqrt{5}\right)^2\)

\(\Leftrightarrow3x=45\)\(\Leftrightarrow x=15\)( thỏa mãn )

Vậy \(x=15\)

c) \(ĐKXĐ:x\inℝ\)

\(\sqrt{x^2+2x+1}=10\)\(\Leftrightarrow\sqrt{\left(x+1\right)^2}=10\)

\(\Leftrightarrow\left|x+1\right|=10\)\(\Leftrightarrow\orbr{\begin{cases}x+1=-10\\x+1=10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-11\\x=9\end{cases}}\)( thỏa mãn )

Vậy \(x=-11\)hoặc \(x=9\)

Check lại đề đi bạn ơi! Chứng minh \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\) thì được