
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


= -2 ( x2 + 2x + 1 ) + 32
= -2 ( x +1 )2 + 32
= - 2 [ (x + 1 )2 - 16 ]
= -2 [ (x + 1 ) 2 - 42 ]
= -2 ( x + 1 - 4 ) ( x + 1 + 4)
= -2 ( x - 3 ) ( x +5 )

\(4x+by+4y+bx=\) \(\left(4x+4y\right)+\left(by+bx\right)\)
\(=\) \(4\left(x+y\right)+b\left(x+y\right)\)
\(=\left(4+b\right)\left(x+y\right)\)
\(2x^2+xy-2x-y=\) \(\left(2x^2+xy\right)-\left(2x+y\right)\)
\(=x\left(2x+y\right)-\left(2x+y\right)\)
\(=\left(x-1\right)\left(2x+y\right)\)

\(16x^4+y^4+4x^2y^2\)
\(=\left(4x^2\right)^2+2.4x^2.y^2+\left(y^2\right)^2-4x^2y^2\)
\(=\left(4x^2+y^2\right)-\left(2xy\right)^2\)
\(=\left(4x^2-2xy+y^2\right)\left(4x^2+2xy+y^2\right)\)
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^3+x^2+4\)
\(=x^3+2x^2-x^2-2x+2x+4\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+2\right)\)

câu a đặt chung x ra là xong
câu b
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé
a) x3 - 2x2 + x - xy2
= x (x2 - 2x + 1 - y2)
= x [(x2 - 2x + 1) - y2]
= x [(x - 1)2 - y2]
= x [(x - 1) + y] [(x - 1) - y]
= x (x - 1 + y) (x - 1 - y)
b) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x3 + 33) - 4x (x + 3)
= (x + 3) (x2 - 3x + 32) - 4x (x + 3)
= (x + 3) [(x2 - 3x + 9) - 4x]
= (x + 3) (x2 - 3x + 9 - 4x)
= (x + 3) (x2 - 7x + 9)
#Học tôt!!!
~NTTH~

Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!

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

\(2x^3+x^2-4x-12=2x^3-4x^2+5x^2-10x+6x-12\)
\(=2x^2\left(x-2\right)+5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+5x+3\right)\)
\(=\left(x-2\right)\left[2x\left(x+1\right)+3\left(x+1\right)\right]\)
\(=\left(x-2\right)\left(x+1\right)\left(2x+3\right)\)
Xin lỗi bạn, mình làm sai.
\(2x^3+x^2-4x-12=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)


Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
\(-2x^2-4x+30\)
\(=-2x^2-10x+6x+30\)
\(=-2x\left(x+5\right)+6\left(x+5\right)\)
\(=\left(-2x+6\right)\left(x+5\right)\)
\(=-2\left(x-3\right)\left(x+5\right)\)
Chúc bạn học tốt.