Question

Tìm a;b;c sao cho:

\(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\le0\)

Answer 1

Vì \(\left(2a+1\right)^2\ge0;\left(b+3\right)^4\ge0;\left(5c-6\right)^4\ge0\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\ge0\)

Mà theo đề bài: \(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\le0\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2=0\)

\(\Rightarrow\begin{cases}\left(2a+1\right)^2=0\\\left(b+3\right)^4=0\\\left(5c-6\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}2a+1=0\\b+3=0\\5c-6=0\end{cases}\)\(\Rightarrow\begin{cases}2a=-1\\b=-3\\5c=6\end{cases}\)\(\Rightarrow\begin{cases}a=\frac{-1}{2}\\b=-3\\c=\frac{6}{5}\end{cases}\)

Vậy \(a=\frac{-1}{2};b=-3;c=\frac{6}{5}\)