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\(x^2-y^2+6x+9=\left(x+3\right)^2-y^2=\left(x+3+y\right)\left(x+3-y\right)\)
\(x^3+3x^2-9x-27=\left(x-3\right)\left(x^2+3x+9\right)+3x\left(x-3\right)=\left(x-3\right)\left(x^2+6x+9\right)=\left(x-3\right)\left(x+3\right)^2\)
A=x4+3x3-9x-9
=x4-9+3x3-9x
=(x2-3)(x2+3)+3x.(x2-3)
=(x2-3)(x2+3+3x)
\(x^2-4y^2+4y-1=x^2-\left(2y-1\right)^2=\left(x+2y-1\right)\left(x-2y+1\right)\)
\(x^4+3x^3-9x-9\)
\(=x^4-9+3x^3-9x\)
\(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3+3x\right)\)
Đặt \(P\left(x\right)=2x^4+3x^3-9x^2-3x+2\)
Giả sử nhân tử của P(x) có dạng : \(P\left(x\right)=2\left(x^2+ax+b\right)\left(x^2+cx+d\right)=\left(x^2+ax+b\right)\left(2x^2+2cx+2d\right)\)
Khai triển : \(P\left(x\right)=2x^4+2cx^3+2dx^2+2ax^3+2acx^2+2adx+2bx^2+2bcx+2bd\)
\(=2x^4+x^3\left(2c+2a\right)+x^2\left(2d+2ac+2b\right)+x\left(2ad+2cb\right)+2bd\)
Dùng phương pháp hệ số bất định :
\(\Rightarrow\begin{cases}2a+2c=3\\2ac+2b+2d=-9\\2ad+2bc=-3\\bd=1\end{cases}\) . Giải ra được \(\begin{cases}a=-1\\b=-1\\c=\frac{5}{2}\\d=-1\end{cases}\)
Vậy \(P\left(x\right)=2\left(x^2-x-1\right)\left(x^2+\frac{5}{2}x-1\right)=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
a) \(3x^2-9x+30=3\left(x^2-3x+10\right)\)
b) \(3x^2-5x-2=3x^2-6x+x-2\)
\(=3x\left(x-2\right)+\left(x-2\right)=\left(3x+1\right)\left(x-2\right)\)
c) \(x^4+4y^4\)
\(=x^4+4y^4+2x^2y^2+2x^2y^2-4x^2y^2+4xy^3-4xy^3+2x^3y-2x^3y\)
\(=\left(4y^4-4xy^3+2x^2y^2\right)+\left(4xy^3-4x^2y^2+2x^3y\right)\)
\(+\left(2x^2y^2-2x^3y+x^4\right)\)
\(=2y^2\left(2y^2-2xy+x^2\right)+2xy\left(2y^2-2xy+x^2\right)\)
\(+x^2\left(2y^2-2xy+x^2\right)\)
\(=\left(2y^2+2xy+x^2\right)\left(2y^2-2xy+x^2\right)\)
d) \(x^5+x+1\)
\(=x^5+x+1+x^4-x^4+x^3-x^3+x^2-x^2\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)^2\left(x+3\right)\)
x3 - 3x2 - 9x + 27
= ( x3 - 3x2 ) - ( 9x - 27 )
= x2( x - 3 ) - 9( x - 3 )
= ( x - 3 )( x2 - 9 )
= ( x - 3 )( x - 3 )( x + 3 )
= ( x - 3 )2( x + 3 )
\(1,9x^3-3x^2+3x-1\)
\(=3x^2.\left(3x-1\right)+\left(3x-1\right)\)
\(=\left(3x^2+1\right).\left(3x-1\right)\)
\(4,x^4-x^3-10x^2+2x+4\)
\(=x^4-3x^3-2x^2+2x^3-6x^2-4x-2x^2-6x-4\)
\(=x^2.\left(x^2-3x-2\right)+2x.\left(x^2-3x-2\right)-2.\left(x^2-3x-2\right)\)
\(=\left(x^2+2x-2\right).\left(x^2-3x-2\right)\)
1/
x2 - 3x - 4
= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)
\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)
\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)
\(=\left(x-4\right)\left(x+1\right)\)
Bài 1 :
\(x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)
nhóm x^4+3x^3 thành 1 nhóm, 9x-27 thành 1 nhóm pn iu
x4+3x3-9x-27
= x(x+3)-9(x+3)
= (x+3)(x-9)