Question

Cho a,b,c và x,y,z khá nhau và khác 0 thỏa mãn :

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\) và \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\). Tính M=\(\sqrt{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}}\)

Answer 1

Vì \(a,b,c,x,y,z\ne0\) nên :

Đặt \(\dfrac{a}{x}=m;\dfrac{b}{y}=n;\dfrac{c}{z}=p\Rightarrow\dfrac{x}{a}=\dfrac{1}{m};\dfrac{y}{b}=\dfrac{1}{n};\dfrac{z}{c}=\dfrac{1}{p}\)

Vậy ta có: \(m+n+p=0\)

\(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}=1\Leftrightarrow\left(\dfrac{1}{m}+\dfrac{1}{m}+\dfrac{1}{p}\right)^2=1\)

\(\Leftrightarrow\dfrac{1}{m^2}+\dfrac{1}{n^2}+\dfrac{1}{p^2}+2\left(\dfrac{m+n+p}{mnp}\right)=1\)

\(\Leftrightarrow\dfrac{1}{m^2}+\dfrac{1}{n^2}+\dfrac{1}{p^2}=1\)

Vậy: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\Rightarrow M=1\)