Bài 1 : Cho $a+b+c=0$ chứng minh rằng


${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=2{{\left( ab+bc+ca \right)}^{2}}$

Bài 2 : Cho$a+b+c=0$; ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ chứng minh rằng ${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=\frac{1}{2}$
Giải
Ta có : $a+b+c=0$

$\Leftrightarrow a=-(b+c)$

$\Leftrightarrow {{a}^{2}}={{b}^{2}}+2bc+{{c}^{2}}$

$\Leftrightarrow {{({{a}^{2}}-{{b}^{2}}-{{c}^{2}})}^{2}}={{\left( 2bc \right)}^{2}}$

$\Leftrightarrow {{a}^{4}}+{{b}^{4}}+{{c}^{4}}-2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}-2{{a}^{2}}{{c}^{2}}=4{{b}^{2}}{{c}^{2}}$

$\Leftrightarrow {{a}^{4}}+{{b}^{4}}+{{c}^{4}}=2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+2{{a}^{2}}{{c}^{2}}$

$\Leftrightarrow 2\left( {{a}^{4}}+{{b}^{4}}+{{c}^{4}} \right)={{a}^{4}}+{{b}^{4}}+{{c}^{4}}+2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+2{{a}^{2}}{{c}^{2}}$

$\Leftrightarrow 2\left( {{a}^{4}}+{{b}^{4}}+{{c}^{4}} \right)={{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}$

$\Leftrightarrow 2\left( {{a}^{4}}+{{b}^{4}}+{{c}^{4}} \right)=1$

$\Leftrightarrow {{a}^{4}}+{{b}^{4}}+{{c}^{4}}=\frac{1}{2}$

Bài 3 : Cho $a,b,c\ne 0$ và $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ . Tính giá trị biểu thức $M=\frac{bc}{{{a}^{2}}}+\frac{ac}{{{b}^{2}}}+\frac{ab}{{{c}^{2}}}$

Bài giải
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\Leftrightarrow \frac{1}{a}+\frac{1}{b}=-\frac{1}{c}$

$\Leftrightarrow {{\left( \frac{1}{a}+\frac{1}{b} \right)}^{3}}={{\left( -\frac{1}{c} \right)}^{3}}$
$\Leftrightarrow \frac{1}{{{a}^{3}}}+3.\frac{1}{{{a}^{2}}b}+3.\frac{1}{a{{b}^{2}}}+\frac{1}{{{b}^{3}}}=-\frac{1}{{{c}^{3}}}$

$\Leftrightarrow \frac{1}{{{a}^{3}}}+\frac{3}{ab}.\left( \frac{1}{a}+\frac{1}{b} \right)+\frac{1}{{{b}^{3}}}=-\frac{1}{{{c}^{3}}}$

$\Leftrightarrow \frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=-\frac{3}{ab}.\left( \frac{1}{a}+\frac{1}{b} \right)=-\frac{3}{ab}.\left( -\frac{1}{c} \right)=\frac{3}{abc}$

$M=\frac{bc}{{{a}^{2}}}+\frac{ac}{{{b}^{2}}}+\frac{ab}{{{c}^{2}}}=\frac{abc}{{{a}^{3}}}+\frac{abc}{{{b}^{3}}}+\frac{abc}{{{c}^{3}}}=abc\left( \frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}} \right)=abc.\frac{3}{abc}=3$

Bài 4 : Cho ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$ và $a,b,c\ne 0$ . Tính $M=\frac{bc}{{{a}^{2}}}+\frac{ac}{{{b}^{2}}}+\frac{ab}{{{c}^{2}}}$
Bài giải

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

$\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\,(ab+bc+ca)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

$\Leftrightarrow ab+bc+ca=0$

$\Leftrightarrow \frac{ab+bc+ca}{abc}=0$

$\Leftrightarrow \frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=0$

$\Leftrightarrow \frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0$

Tương tự bài 3, tính được

$M=\frac{bc}{{{a}^{2}}}+\frac{ac}{{{b}^{2}}}+\frac{ab}{{{c}^{2}}}$= 3