Question

[Nguyen Thi Bich Huong]

Cho x,y,z thỏa mãn đồng thời các hệ thức: \(\left\{{}\begin{matrix}4x^2+4z^2=17\\4y\left(x+2\right)=5\\20y^2+27=-16z\end{matrix}\right.\)

Tính giá trị biểu thức: \(M=10x+4y+2019z\)

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+z^2=\frac{17}{4}\\2xy+4y=\frac{5}{2}\\5y^2+4z=-\frac{27}{4}\end{matrix}\right.\)

\(\Rightarrow x^2+z^2-2xy-4y+5y^2+4z=-5\)

\(\Leftrightarrow\left(x-y\right)^2+\left(z+2\right)^2+\left(2y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\z=-2\\y=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow M=...\)