Bài 1: Cho ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$ và$a,b,c\ne 0$ .

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

Giải

Ta có : ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

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

$\Leftrightarrow ab+bc+ca=0$

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

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

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

Đặt $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$

$\Rightarrow x+y+z=0$ $\Leftrightarrow x+y=-z$$\Leftrightarrow -{{(x+y)}^{3}}={{z}^{3}}$

Lại có : ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$

$={{x}^{3}}+{{y}^{3}}-{{(x+y)}^{3}}$

$={{x}^{3}}+{{y}^{3}}-\left( {{x}^{3}}+3xy(x+y)+{{y}^{3}} \right)$

$={{x}^{3}}+{{y}^{3}}-{{x}^{3}}-{{y}^{3}}-3xy(x+y)$

$=-3xy(x+y)$

$=-3xy.(-z)$

$=3xyz$

Thay $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$ ta được

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$

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

$\Leftrightarrow \frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=\frac{3}{abc}$

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

$\Leftrightarrow \frac{abc}{{{a}^{3}}}+\frac{abc}{{{b}^{3}}}+\frac{abc}{{{c}^{3}}}=3$

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

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

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

Đặt $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$

$\Rightarrow x+y+z=0$ $\Leftrightarrow x+y=-z$$\Leftrightarrow -{{(x+y)}^{3}}={{z}^{3}}$

Lại có : ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$

$={{x}^{3}}+{{y}^{3}}-{{(x+y)}^{3}}$

$={{x}^{3}}+{{y}^{3}}-\left( {{x}^{3}}+3xy(x+y)+{{y}^{3}} \right)$

$={{x}^{3}}+{{y}^{3}}-{{x}^{3}}-{{y}^{3}}-3xy(x+y)$

$=-3xy(x+y)$

$=-3xy.(-z)$

$=3xyz$

Thay $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$ ta được

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$

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

$\Leftrightarrow \frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=\frac{3}{abc}$

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

$\Leftrightarrow \frac{abc}{{{a}^{3}}}+\frac{abc}{{{b}^{3}}}+\frac{abc}{{{c}^{3}}}=3$

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

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

Bài 3: Cho ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ và $a,b,c\ne 0$. Chứng minh rằng

$P=\left( \frac{1}{a}+\frac{1}{b} \right)\left( \frac{1}{b}+\frac{1}{c} \right)\left( \frac{1}{c}+\frac{1}{a} \right)=\frac{8}{abc}$
Giải

${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

$\Leftrightarrow ({{a}^{3}}+{{b}^{3}})+{{c}^{3}}-3abc=0$

$\Leftrightarrow ({{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}})-3ab\left( a+b \right)+{{c}^{3}}-3abc=0$

$\Leftrightarrow {{(a+b)}^{3}}+{{c}^{3}}-3ab\left( a+b \right)-3abc=0$

$\Leftrightarrow \left( a+b+c \right)\left[ {{\left( a+b \right)}^{2}}-(a+b).c+{{c}^{2}} \right]-3ab\left( a+b+c \right)=0$

$\Leftrightarrow \left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+2ab-a.c-bc+{{c}^{2}}-3ab \right]=0$

$\Leftrightarrow \left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ac-bc \right)=0$

$\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ac-bc=0$ vì $a,b,c\ne 0$

$\Leftrightarrow 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2ac-2bc=0$

$\Leftrightarrow ({{a}^{2}}-2ab+{{b}^{2}})+({{b}^{2}}-2bc+{{c}^{2}})+({{c}^{2}}-2ac+{{a}^{2}})=0$

$\Leftrightarrow {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0$

Ta có : ${{(a-b)}^{2}}\ge 0;\,\,{{(b-c)}^{2}}\ge 0;\,\,{{(c-a)}^{2}}\ge 0\,\,\,\forall a,b,c$

Suy ra $\left\{ \begin{align}& {{(a-b)}^{2}}=0 \\ & {{(b-c)}^{2}}=0 \\ & {{(c-a)}^{2}}=0 \\
\end{align} \right.$ $\Leftrightarrow \left\{ \begin{align}& a-b=0 \\ & b-c=0 \\ & c-a=0 \\
\end{align} \right.$$\Leftrightarrow a=b=c$

Vậy $P=\left( \frac{1}{a}+\frac{1}{b} \right)\left( \frac{1}{b}+\frac{1}{c} \right)\left( \frac{1}{c}+\frac{1}{a} \right)=\frac{8}{abc}$

Bài 4: Cho $x,y,z\ne 0$và $x+y+z=xyz$, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}$. Tính $M=\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}$

Giải

Ta có: $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}$

$\Leftrightarrow {{\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)}^{2}}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+2\left( \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \right)=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+\frac{2(x+y+z)}{xyz}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+\frac{2xyz}{xyz}=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}+2=3$

$\Leftrightarrow \frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}=1$