Lời giải:

Nếu $n$ lẻ thì $n+7$ chẵn

$\Rightarrow (n+4)(n+7)$ chẵn 

Nếu $n$ chẵn thì $n+4$ chẵn

$\Rightarrow (n+4)(n+7)$ chẵn 

Vậy $(n+4)(n+7)$ luôn là số chẵn với mọi $n$

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

\(=3^3.3^n+3.3^n+2^3.2^n+2^2.2^n=\)

\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)=30.3^n+12.2^n=\)

\(=6\left(5.3^n+2.2^n\right)⋮6\)

\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+1}\left(9+3\right)+2^{n+2}\left(8+4\right)\)

\(=12.3^{n+1}+12.2^{n+2}=12.\left(3^{n+1}+2^{n+2}\right)\)

mà 12⋮6

\(\Rightarrow12.\left(3^{n+1}+2^{n+2}\right)⋮6\Rightarrow dpcm\)

\(\left(2x-5\right)^2=\left(x-\dfrac{5}{2}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=x-\dfrac{5}{2}\\2x-5=\dfrac{5}{2}-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\3x=\dfrac{15}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)

\(\left(2x-5\right)^2=\left(x-\dfrac{5}{2}\right)^2\)

\(\Leftrightarrow2x-5=x-\dfrac{5}{2}\)

\(\Leftrightarrow2x-x=-\dfrac{5}{2}+5\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

\(...\Rightarrow\dfrac{20\left(x+2\right)}{360}+\dfrac{45\left(x+4\right)}{360}+\dfrac{72\left(x+5\right)}{360}=\dfrac{360\left(x+14\right)}{360}\)

\(\Rightarrow20\left(x+2\right)+45\left(x+4\right)+72\left(x+5\right)=360\left(x+14\right)\)

\(\Rightarrow20x+40+45x+180+72x+360=360x+5040\)

\(\Rightarrow137x+580=360x+5040\)

\(\Rightarrow360x-137x=5040-580\)

\(\Rightarrow223x=4460\Rightarrow x=4460:223=\dfrac{4460}{223}\)

\(\left(-\dfrac{1}{2}\right)^5=-\left(\dfrac{1}{2}\right)^5=-\dfrac{1}{32}\)

\(\left(-\dfrac{2}{3}\right)^4=\left(\dfrac{2}{3}\right)^4=\dfrac{2^3}{3^4}=\dfrac{8}{81}\)