\(2x+6=2\left(x+3\right)\)

\(x^2+9=x^2+9\)

=>MTC sẽ là \(2\cdot\left(x+3\right)\left(x^2+9\right)\)

\(\dfrac{1}{x^2+x}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)};\dfrac{x^2-4}{x^2-1}=\dfrac{x\left(x^2-4\right)}{x\left(x-1\right)\left(x+1\right)}\\ \dfrac{1}{y-1}-\dfrac{1}{y}=\dfrac{y-y+1}{y\left(y-1\right)}=\dfrac{1}{y\left(y-1\right)}\)

ĐKXĐ : 

\(x^4-x^3+2x^2-x+1\ne0\)

\(\Leftrightarrow x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\ne0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+1\right)\ne0\)

Pt

\(\Leftrightarrow x^4+x^3+x+1=0\) 

\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\pm\dfrac{\sqrt{3}\iota+1}{2}\end{matrix}\right.\)

Chắc không cần tìm đkxđ đâu!

MT1: 2x + 2 = 2(x + 1 );  

MT 2: 2x - 2 = 2(x - 1);

1.

Gọi \(d=ƯC\left(2n^2+3n+1;3n+1\right)\)

\(\Rightarrow2n^2+3n+1-\left(3n+1\right)⋮d\)

\(\Rightarrow2n^2⋮d\Rightarrow2n\left(3n+1\right)-3.2n^2⋮d\)

\(\Rightarrow2n⋮d\Rightarrow2\left(3n+1\right)-3.2n⋮d\Rightarrow2⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=2\end{matrix}\right.\)

\(d=2\Rightarrow3n+1=2k\Rightarrow n=2m+1\)

\(\Rightarrow n\) lẻ thì A không tối giản

\(\Rightarrow n\) chẵn thì A tối giản

2.

Giả thiết tương đương:

\(xy^2+\dfrac{x^2}{z}+\dfrac{y}{z^2}=3\)

Đặt \(\left(x;y;\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a^2c+b^2a+c^2b=3\)

Ta có: \(9=\left(a^2c+b^2a+c^2b\right)^2\le\left(a^4+b^4+c^4\right)\left(c^2+a^2+b^2\right)\)

\(\Rightarrow9\le\left(a^4+b^4+c^4\right)\sqrt{3\left(a^4+b^4+c^4\right)}\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)^3\ge81\Rightarrow a^4+b^4+c^4\ge3\)

\(\Rightarrow M=\dfrac{1}{a^4+b^4+c^4}\le\dfrac{1}{3}\)

\(M_{max}=\dfrac{1}{3}\) khi \(\left(a;b;c\right)=\left(1;1;1\right)\) hay \(\left(x;y;z\right)=\left(1;1;1\right)\)

\(\dfrac{x+y}{2-x}=\dfrac{-\left(x+y\right)}{x-2}\)

\(\dfrac{-y}{y-4}=\dfrac{--y}{4-y}=\dfrac{y}{4-y}\)

ĐKXĐ: \(x\ne1\)

Ta có: \(B=\dfrac{x^4-2x^3-3x^2+8x-1}{x^2-2x+1}\)

\(=\dfrac{x^4-2x^3+x^2-4x^2+8x-4+3}{x^2-2x+1}\)

\(=\dfrac{x^2\left(x^2-2x+1\right)-4\left(x^2-2x+1\right)+3}{x^2-2x+1}\)

\(=\dfrac{\left(x-1\right)^2\cdot\left(x^2-4\right)+3}{\left(x-1\right)^2}\)

\(=x^2-4+\dfrac{3}{\left(x-1\right)^2}\)

Để B nguyên thì \(3⋮\left(x-1\right)^2\)

\(\Leftrightarrow\left(x-1\right)^2\inƯ\left(3\right)\)

\(\Leftrightarrow\left(x-1\right)^2\in\left\{1;3;-1;-3\right\}\)

mà \(\left(x-1\right)^2>0\forall x\) thỏa mãn ĐKXĐ

nên \(\left(x-1\right)^2\in\left\{1;3\right\}\)

\(\Leftrightarrow x-1\in\left\{1;9\right\}\)

hay \(x\in\left\{2;10\right\}\) (nhận)

Vậy: \(x\in\left\{2;10\right\}\)