Question

Cho a,b,c ≠0, 4a+5b-6c=-5, \(\frac{1}{4a}\)+\(\frac{1}{5b}\)-\(\frac{1}{6c}\)=0

Tính 16a²+25b²+36c²

Answer 1

Đặt \(\left(4a;5b;-6c\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=-5\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=25\\\frac{xy+yz+zx}{xyz}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+zx\right)=25\\xy+yz+zx=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2=25\) hay \(16a^2+25b^2+36c^2=25\)