\(x+22⋮x+1\)

\(\Rightarrow x+1+21⋮x+1\)

     \(x+1⋮x+1\)

\(\Rightarrow21⋮x+1\)

\(\Rightarrow x+1\inƯ\left(21\right)\)

\(\Rightarrow x+1\in\left\{-1;1;-3;3;-7;7;-21;21\right\}\)

\(\Rightarrow x\in\left\{-2;0;-4;2;-7;6;-22;20\right\}\)

a) (2n+1) chia hết (6-n)

=>(2n+1)+(12-2n) chia hết cho (6-n)

=>13 chia hết (6-n)

=> 6-n thuộc ước của 13={ 1;-1;13;-13}

Nếu 6-n=1=>n=5

Nếu 6-n=-1=>n=7

Nếu 6-n=13=> n=-7 (loại)

Nếu 6-n=-13=>n=19

Vậy n=5;n=7;n=19

N

Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

\(=\left(y^2-2y+1\right)-x^2\)

\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)

c: \(\left(x+1\right)^2-25\)

=(x+1+5)(x+1-5)

=(x-4)(x+6)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)