Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)

\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Áp dụng vào bài toán ta có: 

\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

.....

\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)

\(\hept{\begin{cases}x^2-2y^2=2x+y\left(1\right)\\y^2-2x^2=2y+x\left(2\right)\end{cases}}\)

\(\left(1\right)-\left(2\right)\)

\(\Leftrightarrow3x^2-3y^2=x-y\)

\(\Leftrightarrow\left(x-y\right)\left(3x+3y-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\3x+3y-1=0\end{cases}}\)

TH1: x=y => x² - 2x² =2x+x => -x² - 3x=0 

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Th2: (làm tương tự TH1)