\(2x^4+x^3-6x^2+x+2\) 

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

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

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

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

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

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

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

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

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

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

\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)

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

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

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

\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

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

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

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

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

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

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

x(2x^2-x-6)

x(2x^2-4x+3x-6)

x[2x(x-2)+3(x-2)]

x(2x+3)(x-2)

2x^5-6x^4-2a^2x^3-6ax^3

 =(2x^5-2a^2x^3)-(6x^4+6ax^3)

 =2x^3(x^2-a^2)-6x^3(x+a)

 =2x^3(x-a)(x+a)-6x^3(x+a)

 =(x+a)(2x^4-2x^3a-6x^3)

 =(x+a) 2x^3 (x-a-3)

đặt y=x²+1

=>y²=(x²+1)²

y²=x⁴+2x²+1

đặt P(x)=x^4+6x^3+11x^2+6x+1

=x⁴+2x²+1+6x³+6x+9x²

=x⁴+2x+1+6x(x²+1)+9x²

thay y²=x⁴+2x²+1 và y=x²+1 ta được 

Q(y)=y²+6xy+9x²

=(y+3x)²

thay y=x²+1 ta được:

(x²+3x+1)²

vậy x^4+6x^3+11x^2+6x+1=(x²+3x+1)²

giup e với

      \(x^4+6x^3+7x^2-6x+1\)

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2\left(x^2+3x\right)+1\)

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

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

Chúc bạn học tốt.

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

\(=x^4+6x^3+9x^2-2x^2-6x+1\)

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

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

https://olm.vn/hoi-dap/tim-kiem?q=Cho+%CE%94ABC+vu%C3%B4ng+t%E1%BA%A1i+A+c%C3%B3+AB+%3E+AC,+M+l%C3%A0+%C4%91i%E1%BB%83m+tu%E1%BB%B3+%C3%BD+tr%C3%AAn+BC.+Qua+M+k%E1%BA%BB+Mx+vu%C3%B4ng+g%C3%B3c+v%E1%BB%9Bi+BC+v%C3%A0+c%E1%BA%AFt+AB+t%E1%BA%A1i+I+c%E1%BA%AFt+CA+t%E1%BA%A1i+D.a.+Ch%E1%BB%A9ng+minh+%CE%94ABC+%C4%91%E1%BB%93ng+d%E1%BA%A1ng+v%E1%BB%9Bi+%CE%94MDCb.+Ch%E1%BB%A9ng+minh:+BI.BA+=+BM.BCc.+Cho+g%C3%B3c+ACB+=+60o+v%C3%A0+S%CE%94CDB+=+60+cm2.+T%C3%ADnh+S%CE%94CMAGi%C3%BAp+m%C3%ACnh+c%C3%A2u+c+v%E1%BB%9Bi&id=573451

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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