Question

Cho \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)và \(x;y;z\ne0\)

CMR \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{1}{a^2+b^2+c^2}\)

Answer 1

Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=k\)

\(\Rightarrow\hept{\begin{cases}a=kx;b=ky;c=kz\Rightarrow a^2=k^2x^2;b^2=k^2y^2;c^2=k^2z^2\\a+b+c=k\left(x+y+z\right)\end{cases}}\)

Có: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{x^2+y^2+z^2}{\left(kx^2+ky^2+kz^2\right)^2}=\frac{x^2+y^2+z^2}{k^2\left(x^2+y^2+z^2\right)^2}=\frac{1}{k^2\left(x^2+y^2+z^2\right)}\)

\(=\frac{1}{k^2x^2+k^2y^2+k^2z^2}=\frac{1}{a^2+b^2+c^2}\)(đpcm)