Question

Cho x, y, z là 3 số khác 0 thỏa mãn : \(\left\{{}\begin{matrix}x+2y+3z=4\\\dfrac{1}{x}+\dfrac{1}{2y}+\dfrac{1}{3z}=0\end{matrix}\right.\)

Tính P = \(4y^2+x^2+9z^2\)

Answer 1

Lời giải:

Ta có:

\(\left\{\begin{matrix} x+2y+3z=4\\ \frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ \frac{6yz+2xy+3xz}{6xyz}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+2y+3z=4\\ 2xy+6yz+3xz=0\end{matrix}\right.\)

Do đó:

\((x+2y+3z)^2-2(2xy+6yz+3xz)=4^2-2.0=16\)

\(\Leftrightarrow x^2+4y^2+9z^2=16\)

\(\Leftrightarrow P=16\)