Ta có:

\(x^3+2x^2+x+2\)

\(=x^2\left(x+2\right)+\left(x+2\right)\)

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

x³+2x²+x+2=x²(x+2)+(x+2)

                  =(x²+1)(x+2)

cam on ban nhieu lam

\(a.x^3+3x^2+4x+2\)

\(=x^3+x^2+2x^2+2x+2\)

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

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

\(b.6x^4-x^3-7x^2+x+1\)

\(=6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1\)

\(=6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(6x^3+5x^2-2x-1\right)\)

\(=\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)\)

\(=\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left[3x\left(2x-1\right)+\left(2x-1\right)\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)\)

k giùm cái cho đỡ buồn!

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)

b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)

c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)

d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)

a) 

/ \(x^4+x^2-2=0\)

\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

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

\(=\left(3x-9\right)^2-\left(2x+2\right)^2\)

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

c: =(x-3-2y)(x-3+2y)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

\(x^2-4x-y^2+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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