b: Ta có: \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6^2=36\)
=>\(a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=36\)
=>\(14+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=36\)
=>\(2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=36-14=22\)
=>\(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\frac{22}{2}=11\)
\(a+11=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
\(=\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
\(b+11=b+\sqrt{ba}+\sqrt{bc}+\sqrt{ac}\)
\(=\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)+\sqrt{c}\left(\sqrt{b}+\sqrt{a}\right)=\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
c+11
\(=c+\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\)
\(=\sqrt{c}\left(\sqrt{c}+\sqrt{b}\right)+\sqrt{a}\left(\sqrt{c}+\sqrt{b}\right)=\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
Ta có: \(\frac{\sqrt{a}}{a+11}+\frac{\sqrt{b}}{b+11}+\frac{\sqrt{c}}{c+11}\)
\(=\frac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\frac{\sqrt{b}}{\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\frac{\sqrt{c}}{\left(\sqrt{c}+\sqrt{a}\right)\left(\sqrt{c}+\sqrt{b}\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\frac{22}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\) (1)
Ta có: \(\frac{22}{\sqrt{\left(a+11\right)\left(b+11\right)\left(c+11\right)}}\)
\(=\frac{22}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{a}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{c}+\sqrt{a}\right)\left(\sqrt{c}+\sqrt{b}\right)}}\)
\(=\frac{22}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\cdot\left(\sqrt{b}+\sqrt{c}\right)^2\cdot\left(\sqrt{a}+\sqrt{c}\right)^2}}=\frac{22}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\) (2)
Từ (1),(2) suy ra \(\frac{\sqrt{a}}{a+11}+\frac{\sqrt{b}}{b+11}+\frac{\sqrt{c}}{c+11}=\frac{22}{\sqrt{\left(a+11\right)\left(b+11\right)\left(c+11\right)}}\)
a: \(x^2+x+5\sqrt{x^2-1}\)
\(=x\left(x+1\right)+5\cdot\sqrt{x+1}\cdot\sqrt{x-1}\)
\(=\sqrt{x+1}\left(x\cdot\sqrt{x+1}+5\cdot\sqrt{x-1}\right)\)
\(5x-5+x\cdot\sqrt{x^2-1}\)
\(=5\cdot\left(x-1\right)+x\cdot\sqrt{\left(x-1\right)\left(x+1\right)}\)
\(=\sqrt{x-1}\left(x\cdot\sqrt{x+1}+5\cdot\sqrt{x-1}\right)\)
Ta có: \(M=\frac{x^2+x+5\sqrt{x^2-1}}{5x-5+x\cdot\sqrt{x^2-1}}\)
\(=\frac{\sqrt{x+1}\left(x\cdot\sqrt{x+1}+5\cdot\sqrt{x-1}\right)}{\sqrt{x-1}\left(x\cdot\sqrt{x+1}+5\cdot\sqrt{x-1}\right)}=\frac{\sqrt{x+1}}{\sqrt{x-1}}=\sqrt{\frac{x+1}{x-1}}\)
Khi \(x=-1-\sqrt2\) thì \(M=\sqrt{\frac{-1-\sqrt2+1}{-1-\sqrt2-1}}=\sqrt{\frac{-\sqrt2}{-2-\sqrt2}}=\sqrt{\frac{1}{\sqrt2+1}}=\sqrt{\sqrt2-1}\)
