Câu hỏi của Law Trafargal

Question

Cho a,b thỏa mãn :10a^2+5b^2+12ab+4a-6b+13=0
. Tính giá trị biểu M=(2a+b)^2020

Nguyễn Việt Lâm · Accepted Answer

$9a^2+12ab+4b^2+a^2+4a+4+b^2-6b+9=0$

$\Leftrightarrow\left(3a+2b\right)^2+\left(a+2\right)^2+\left(b-3\right)^2=0$

$\Rightarrow\left\{{}\begin{matrix}3a+2b=0\a+2=0\b-3=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}a=-2\b=3\end{matrix}\right.$

$M=\left(-4+3\right)^{2020}=1$Câu trả lời: $9a^2+12ab+4b^2+a^2+4a+4+b^2-6b+9=0$

$\Leftrightarrow\left(3a+2b\right)^2+\left(a+2\right)^2+\left(b-3\right)^2=0$

$\Rightarrow\left\{{}\begin{matrix}3a+2b=0\a+2=0\b-3=0\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}a=-2\b=3\end{matrix}\right.$

$M=\left(-4+3\right)^{2020}=1$