Phân tích đa thức thành nhân tử:
x3 - x2 - 3x2 + 6x - 3
Giúp tôi với >__< ~
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
a: =3x^2-3x-8x+8=(x-1)(3x-8)
b: =x^2-x-5x+5=(x-1)(x-5)
c: =x^2-6x+2x-12=(x-6)(x+2)
a) \(4x^3y^2-8x^2y+12xy^2=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x-2y\right)^2-49=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-6x+9\right)-25=\left(x-3\right)^2-25=\left(x-3-5\right)\left(x-3+5\right)=\left(x-8\right)\left(x+2\right)\)
a) 4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)4x3y2−8x2y+12xy2=4xy(x2y−2x+3y)
b) 3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)3x2−6xy−5x+10y=3x(x−2y)−5(x−2y)=(x−2y)(3x−5)
c) x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)x2−49+4y2−4xy=(x−2y)2−49=(x−2y−7)(x−2y+7)
d) x2−6x−16=(x2−6x+9)−25=(x−3)2−25=(x−3−5)(x−3+5)=(x−8)(x+2)
a) \(4x^3y^2-8x^2y+12xy^2=4xy.x^2y-4xy.2x+4xy.3y=4xy\left(x^2y-2x+3y\right)\)
b) \(3x^2-6xy-5x+10y=\left(3x^2-6xy\right)-\left(5x-10y\right)=3x\left(x-2y\right)-5\left(x-2y\right)=\left(x-2y\right)\left(3x-5\right)\)
c) \(x^2-49+4y^2-4xy=\left(x^2-4xy+4y^2\right)-49=\left(x-2y\right)^2-7^2=\left(x-2y-7\right)\left(x-2y+7\right)\)
d) \(x^2-6x-16=\left(x^2-8x\right)+\left(2x-16\right)=x\left(x-8\right)+2\left(x-8\right)=\left(x-8\right)\left(x+2\right)\)
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
Bài 1:
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)
1, =3x (2x -3y)
c, = 3x(x-y) -2(x-y)
= (3x-2)(x-y)
2, Ta có: x2 -6x+10= (x-3)2 +11
Nhận xét: (x-3)2 >= 0 với mọi số thực x
=> (x-3)2 +1 >= 1 >0 (đpcm)
\(=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
Ta có: \(x^3-\left(y-2\right)^3+\left(y-x-2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)+\left(x-y+2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4+x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+y^2+6+xy-x-5y\right)\)
x3 - x2 - 3x2 + 6x - 3
= x3 - x2 - 3x2 + 3x + 3x - 3
= x2 ( x - 1 ) - 3x ( x - 1 ) + 3 ( x - 1 )
= ( x - 1 ) ( x2 - 3x + 3 )