Áp dụng BĐT Cô-si cho 2 số dương \(9b,\left(a+8b\right)\)ta có:

\(\frac{9b+a+8b}{2}\ge\sqrt{9b\left(a+8b\right)}\Rightarrow a\sqrt{9b\left(a+8b\right)}\le\frac{a^2+17ab}{2}\)

Tương tự có: \(b\sqrt{9a\left(b+8a\right)}\le\frac{b^2+17ab}{2}\)

Nên: \(M=a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\le\frac{a^2+b^2+34ab}{2}\)

Mà \(2ab\le a^2+b^2\Rightarrow M\le\frac{18\left(a^2+b^2\right)}{2}\Rightarrow M\le9.16=144\)

Dấu ''='' xảy ra \(\Leftrightarrow a=b=2\sqrt{2}\)

Đã học đến cấp 2 đâu mà làm 

\(\frac{1-x}{\sqrt{x}}=2\)   

\(ĐK:x>0\)   

\(1-x=2\sqrt{x}\)    

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

\(1-2x+x^2=4x\)   

\(x^2-6x+1=0\)    

\(\orbr{\begin{cases}x=3+2\sqrt{2}\\x=3-2\sqrt{2}\end{cases}}\)

\(P=\frac{1-x}{\sqrt{x}}=2\)ĐK : \(x\ge0\)

\(\Leftrightarrow1-x=2\sqrt{x}\)bình phương 2 vế ta được : 

\(\Leftrightarrow\left(x-1\right)^2=4x\Leftrightarrow x^2-2x+1=4x\)

\(\Leftrightarrow x^2-6x+1=0\)Ta có : 

\(\Delta=\left(-6\right)^2-4.1=36-4=32>0\)

\(x_1=\frac{6-\sqrt{32}}{2};x_2=\frac{6+\sqrt{32}}{2}\)

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

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

Phương trình tương đương : 

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

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