Question

Tìm a,b,c biết

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

b,\(\left(a-7\right)^2+\left(3b+2\right)^2+\left(4c-5\right)^6< =0\)

c,\(\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+19\right)^6< =0\)

d,\(\left(7b-3\right)^4+\left(21a-6\right)^4+\left(18c+5\right)^6< =0\)

Answer 1

a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)

\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)

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

Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)

- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;