Bài 1:
a) \(\left(n+1\right)^3=2\left(n+1\right)^2\)

\(\Rightarrow\left(n+1\right)^3-2\left(n+1\right)^2=0\)

\(\Rightarrow\left(n+1\right)^2\left[\left(n+1\right)-2\right]\)

\(\Rightarrow\left(n+1\right)^2\left(n-1\right)=0\)

\(\Rightarrow\left\{\begin{matrix}\left(n+1\right)^2=0\-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n+1=0\=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n=-1\=1\end{matrix}\right.\)

Vậy...

b) Ta có: \(3n-7⋮6n+5\)

\(\Rightarrow6n-14⋮6n+5\)

\(\Rightarrow\left(6n+5\right)-19⋮6n+5\)

\(\Rightarrow19⋮6n+5\)

\(\Rightarrow6n+5\in\left\{1;-1;19;-19\right\}\)

\(\left\{\begin{matrix}6n+5=1\\6n+5=-1\\6n+5=19\\6n+5=-19\end{matrix}\right.\Rightarrow\left\{\begin{matrix}n=-\frac{2}{3}\left(loại\right)\=-1\left(t.m\right)\=\frac{7}{3}\left(loại\right)\=-4\left(t.m\right)\end{matrix}\right.\)

Vậy \(n\in\left\{-1;-4\right\}\)

Bài 5:

\(xy+4x+3y+12=7\)

\(\Rightarrow x\left(y+4\right)+3\left(y+4\right)=7\)

\(\Rightarrow\left(x+3\right)\left(y+4\right)=7\)

\(\Rightarrow\left\{\begin{matrix}x+3=7\\y+4=1\end{matrix}\right.\) hoặc \(\left\{\begin{matrix}x+3=-7\\y+4=-1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=4\\y=-3\end{matrix}\right.\) hoặc \(\left\{\begin{matrix}x=-10\\y=-5\end{matrix}\right.\)

Vậy...

Bài 7:

a) Ta có: \(\left(2x-6\right)^{10}+\left|3x+y\right|\ge0\)

\(\Rightarrow A=\left(2x-6\right)^{10}+\left|3x+y\right|-7\ge-7\)

\(\Rightarrow MIN_A=-7\) khi \(\left\{\begin{matrix}\left(2x-6\right)^{10}=0\\\left|3x+y\right|=0\end{matrix}\right.\)

Vậy \(MIN_A=-7\) khi \(x=3;y=-9\)