\(9^{9^{9^9}}-9^{9^9}=9^{2a+1}-9^{2b+1}\equiv9-9\equiv0\left(mod10\right)\)

\(\left(x+1\right)\left(y+1\right)=2\)

\(\Leftrightarrow x=\frac{1-y}{1+y}\)

\(P=\sqrt{x^2+y^2-\sqrt{2\left(x^2+1\right)\left(y^2+1\right)}+2}+xy\)

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

\(=\sqrt{\left(\frac{1-y}{1+y}\right)^2+y^2-2.\frac{y^2+1}{y+1}+2}+\left(\frac{1-y}{1+y}\right)y\)

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

\(=\frac{y^2+1}{y+1}+\left(\frac{1-y}{1+y}\right)y=1\) 

tận cùng là 6

\(\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}\)

\(=\)\(\sqrt{\left(\sqrt{a+b}\right)^2+2\sqrt{a+b}\sqrt{c}+\left(\sqrt{c}\right)^2}+\sqrt{\left(\sqrt{a+b}\right)^2-2\sqrt{a+b}\sqrt{c}+\left(\sqrt{c}\right)^2}\)

\(=\)\(\sqrt{\left(a+b+c\right)^2}+\sqrt{\left(a+b-c\right)^2}\)

\(=\)\(\left|a+b+c\right|+\left|a+b-c\right|\)

Đến đây e ko bít làm tiếp -_- 

Chúc chị học tốt ~ 

Đề z thì c/m kiểu gì bn ei

Ta có:\(a+\left(b+c\right)\ge2\sqrt{a\left(b+c\right)}\Rightarrow\frac{a+b+c}{2}\ge\sqrt{a\left(b+c\right)}\)

 \(\Rightarrow\sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a\left(b+c\right)}}\ge\frac{a}{\frac{a+b+c}{2}}=\frac{2a}{a+b+c}\)
Chứng minh tương tự rồi cộng vế với vế ta được:
\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge\frac{2\left(a+b+c\right)}{a+b+c}=2\)          
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=b+c\\b=c+a\\c=a+b\end{cases}}\)-> hệ vô nghiệm
\(\)\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2\)

Ta có đpcm