a, \(\left(2x-1\right)^2-\left(3x-1\right)^2=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=-x\left(5x-2\right)\)

b, \(\left(x+1\right)^2-9=\left(x+1-3\right)\left(x+1+3\right)=\left(x-2\right)\left(x+4\right)\)

c, \(\left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\)

d, \(x^2-9=\left(x-3\right)\left(x+3\right)\); e, \(x^2-25=\left(x-5\right)\left(x+5\right)\)

f, \(\left(x+2\right)^2-\left(3x-1\right)^2=\left(x+2-3x+1\right)\left(x+2+3x-1\right)=\left(-2x+3\right)\left(4x+1\right)\)

i, \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

a, \(P=\left(\frac{x+2}{x-2}+\frac{x}{x+2}-\frac{4}{x^2-4}\right):\left(x-\frac{x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{x^2+4x+4+x^2-2x-4}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2+2x-x^2}{x+2}\right)\)

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

b, Ta có : \(\frac{x+1}{x-2}=\frac{x-2+3}{x-2}=1+\frac{3}{x-2}\)

\(\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Thay từng giá trị vào P ta được : x = 3 ; -1 ; 5 thì P nhận giá trị nguyên dương 

c, Ta có : \(-x^2-x-2=0\Leftrightarrow-\left(x^2+x+2\right)=0\)

Vì \(x^2+x+2=x^2+x+\frac{1}{4}+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)

nên ko có giá trị x thỏa mãn 

d, Ta có : \(P=\frac{x+1}{x-2}>0\)

TH1 : \(\hept{\begin{cases}x+1>0\\x-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-1\\x>2\end{cases}}\Leftrightarrow x>2\)

TH2 : \(\hept{\begin{cases}x+1< 0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x< 2\end{cases}}\Leftrightarrow x< -1}\)

Vậy x > 2 ; x < -1 thì P > 0 

Ta có |x - 2| = |2x - 1| 

<=> \(\orbr{\begin{cases}x-2=2x-1\\x-2=-2x+1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy tập nghiệm phương trình là : S = \(\left\{1;-1\right\}\)

10 nha