Vì cả 3 số hạng đều có mũ chẵn và tổng bằng 0

=>(2a+1)^2=0

=>2a+1=0

=>2a=-1

=>a=-0,5

=>b=-3

=>c=1,2

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

Vì (2a+1)² >=0;(b+3)^4>=0;(5c-6)² >=0

\(\Rightarrow\)Không tìm được a,b,c

Ta có: \(\left(2a+1\right)^2\ge0,\left(b+3\right)^4\ge0,\left(5c-6\right)^2\ge0\), mọi a, b, c

=> \(\left(2a+1\right)^2+\left(b+3\right)^4+\left(5c-6\right)^2\ge0\), mọi a, b, c

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

Vì thế chỉ có thể xảy ra là dấu bằng 

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

<=> 2a+1=0, b+3=0, 5c-6=0

<=> a=-1/2, b=-3, c=6/5

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

Do \(\left(2a+1\right)^2\ge0\)

\(\left(b+3\right)^4\ge0\)

\(\left(5c-6\right)^2\ge0\)

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

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

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

*) \(\left(2a+1\right)^2=0\)

\(\Rightarrow2a+1=0\)

\(2a=-1\)

\(a=-\dfrac{1}{2}\)

*) \(\left(b+3\right)^4=0\)

\(\Rightarrow b+3=0\)

\(b=-3\)

*) \(\left(5c-6\right)^2=0\)

\(\Rightarrow5c-6=0\)

\(5c=6\)

\(c=\dfrac{6}{5}\)

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

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

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 ;-;