Ta có: 

\(\hept{\begin{cases}\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}\\\left(\sqrt{2}-\sqrt{3}\right)^2=5-2\sqrt{6}\end{cases}}\)

Ta chứng minh: Với mọi \(n\in N;n>0\)thì \(\left(5+2\sqrt{6}\right)^n+\left(5-2\sqrt{6}\right)^n\in Z\)

Với \(n=1\)thì \(\left(5+2\sqrt{6}\right)^1+\left(5-2\sqrt{6}\right)^1=10\in Z\)

Với \(n=2\)thì \(\left(5+2\sqrt{6}\right)^2+\left(5-2\sqrt{6}\right)^2=98\in Z\)

Giả sử nó đúng đến \(n=k\)hay

\(\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right)^k=a\in Z\)

Ta chứng minh nó đúng với \(n=k+1\) hay \(\hept{\begin{cases}\left(5+2\sqrt{6}\right)^{k-1}+\left(5-2\sqrt{6}\right)^{k-1}=a\in Z\\\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right)^k=b\in Z\end{cases}}\)

Ta có:

\(\left(5+2\sqrt{6}\right)^{k+1}+\left(5-2\sqrt{6}\right)^{k+1}\) \(=\left(5+2\sqrt{6}\right).\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^k\)

\(=\left(5+2\sqrt{6}\right).\left(b-\left(5-2\sqrt{6}\right)^k\right)+\left(5-2\sqrt{6}\right).\left(b-\left(5+2\sqrt{6}\right)^k\right)\)

\(=b\left(\left(5+2\sqrt{6}\right)+\left(5-2\sqrt{6}\right)\right)-\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^k-\left(5-2\sqrt{6}\right).\left(5+2\sqrt{6}\right)^k\)

\(=10b-\left(5-2\sqrt{6}\right)^{k-1}-\left(5+2\sqrt{6}\right)^{k-1}\)

\(=10b-a\in Z\)

Vậy theo quy nạp thì nó đúng.

Quay lại bài toán thì ta có:

\(\left(\sqrt{2}+\sqrt{3}\right)^{2310}+\left(\sqrt{2}-\sqrt{3}\right)^{2310}=\left(5+2\sqrt{6}\right)^{1155}+\left(5-2\sqrt{6}\right)^{1155}\in Z\)

Đề bị thiếu rồi. Đáng lẽ phải có a + b = ??? đấy nữa chứ.

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

\(\Rightarrow\sqrt{x^2+3}-x=\sqrt{y^2+3}+y\)(1)

ttu \(\sqrt{y^2+3}-y=\sqrt{x^2+3}+x\) (2)

lay (1)+(2)

\(-\left(x+y\right)=x+y\Rightarrow x+y=0\)

ma \(A=x^{2013}+y^{2013}+1=\left(x+y\right)M+1=1\)

???????????

\(x^2+9x+20=2\sqrt{3x+10}\\ \)

\(x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\\ \)

\(\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\\ \)

=> \(\hept{\begin{cases}x+3=0\\3x+9=0\end{cases}=>x=-3}\)

Giải kiểu này nhanh gọn hơn.

Giải:

Ta có: 

\(x^2+9x+20=2\sqrt{3x+10}\)

\(\Leftrightarrow\sqrt{3x+10}-1^2+x+3^2=0\)

\(\Leftrightarrow x=-3\)

Ta chứng minh bất đẳng thức phụ

\(\frac{1}{8x^2+1}\ge\frac{2}{x+1}-1\)

\(\Leftrightarrow4x^3-4x^2+x\ge0\)

\(\Leftrightarrow x\left(2x-1\right)^2\ge0\)(đúng)

Áp dụng vào bài toán ta được

\(\frac{1}{8a^2+1}+\frac{1}{8b^2+1}+\frac{1}{8c^2+1}\ge-1+\frac{2}{a+1}-1+\frac{2}{b+1}-1+\frac{2}{c+1}\)

\(=-3+2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)=-3+4=1\)

Ta có: \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\)

\(\Rightarrow3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)=1\)

\(\Rightarrow\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1\)

Xét BĐT  \(\Sigma_{cyc}\frac{1}{8a^2+1}\ge1\Leftrightarrow3-\Sigma_{cyc}\frac{1}{8a^2+1}\le2\)

\(\Leftrightarrow\Sigma_{cyc}\frac{8a^2}{8a^2+1}\le2\Leftrightarrow\Sigma_{cyc}\frac{4a^2}{8a^2+1}\le2\)

Xét BĐT phụ: \(\frac{4x^2}{8x^2+1}\le\frac{x}{x+1}\)(*)

Thật vậy: (*)\(\Leftrightarrow\frac{x\left(2x-1\right)^2}{\left(x+1\right)\left(8x^2+1\right)}\)(đúng với mọi x thực dương)

Áp dụng, ta có: \(\Sigma_{cyc}\frac{4a^2}{8a^2+1}\le\text{​​}\Sigma_{cyc}\frac{a}{a+1}=1\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)