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 ;-;

ok

tất cả đều mũ chẳn nên lớn hơn hoặc bằng 0 => để thõa mãn các tổng cộng lại bằng 0 => mỗi tổng bằng 0 

a, Vì \(\hept{\begin{cases}\left(12a-9\right)^2\ge0\\\left(8b+1\right)^4\ge0\\\left(c+15\right)^6\ge0\end{cases}\Rightarrow\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+15\right)^6\ge0}\)

Mà \(\left(12a-9\right)^2+\left(8b+1\right)^4+\left(c+15\right)^6\le0\)

\(\Rightarrow\hept{\begin{cases}\left(12a-9\right)^2=0\\\left(8b+1\right)^4=0\\\left(c+15\right)^6=0\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{3}{4}\\b=\frac{-1}{8}\\c=-15\end{cases}}}\)

b, tương tự a

Lời giải:
a. 

$f(-1)=a-b+c$

$f(-4)=16a-4b+c$

$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$

$\Rightarrow f(-4)=6f(-1)$

$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)

b.

$f(-2)=4a-2b+c$

$f(3)=9a+3b+c$

$\Rightarrow f(-2)+f(3)=13a+b+2c=0$

$\Rightarrow f(-2)=-f(3)$

$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)

a. 

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nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)