Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.5x2-10xy+5y2-20z2
=5(x2-2xy+y2-4z2)
=5[ (x2-2xy+y2)-(2z)2 ]
=5[ (x-y)2-(2z)2 ]
=5(x-y-2z)(x-y+2z)
b.16x-5x2-3
=15x+x-5x2-3
=(15x-3)+(x-5x2)
=3(5x-1)+x(1-5x)
=3(5x-1)-x(5x-1)
=(5x-1)(3-x)
c.x2-5x+5y-y2
=(5y-5x)+(x2-y2)
=5(y-x)+(x-y)(x+y)
=5(y-x)-(y-x)(y+x)
=(y-x)[5-(y+x)]
=(y-x)(5-y-x)
d.3x2-6xy+3y2-12z2 (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)
=3(x2-2xy+y2-4z2)
=3[ (x2-2xy+y2)-(2z)2 ]
=3[ (x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
e.x2+4x+3
=x2+3x+x+3
=(x2+x)+(3x+3)
=x(x+1)+3(x+1)
=(x+1)(x+3)
f.(x2+1)2-4x2
=(x2+1)2-(2x)2
=(x2+1-2x)(x2+1+2x)
h.x2-4x-5
=x2-5x+x-5
=(x2+x)+(-5x-5)
=x(x+1)-5(x+1)
-(x+1)(x-5)
1)\(x^4+2x^3+x^2\)
=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra
=\(x^2\left(x+1\right)^2\)
2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
=\(\left(x+y\right)^3-\left(x+y\right)\)
=\(\left(x+y\right)\left(\left(x+y\right)^2+1\right)\)
3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)
=\(\left(x+y\right)^2-4z^2\)
=\(\left(x+y+2z\right)\left(x+y-2z\right)\)
4)pt => \(2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
=\(2\left(x-y\right)-\left(x-y\right)^2\)
=\(\left(x-y\right)\left(2-x+y\right)\)
k chi nha
a, x4 + 2x3 +x2 = x4 +x3 +x3 +x2 =(x4+x3 )+(x3 +x2 ) =x3(x +1 ) + x2 (x+1 ) =(x+1)(x3+x2)
a) x4 + 2x3 + x2
= x2(x2 + 2x + 1)
= x2(x + 1)2
= [x(x + 1)]2
= (x2 + x)2
b) 5x3 - 10xy + 5y2 - 20z2
= 5(x3 - 2xy + y2 - 4z2)
c) x2y - xy2 + x3 - y3
= xy(x - y) + (x - y)(x2 + xy + y2)
= (x - y)(x2 + 2xy + y2)
= (x - y)(x + y)2
d) x2 - xy + 4x - 2y + 4
= (x2 + 4x + 4) - (xy + 2y)
= (x + 2)2 - y(x + 2)
= (x + 2)(x + 2 - y)
d) x2 - x - 6
= x2 - 3x + 2x - 6
= x(x - 3) + 2(x - 3)
= (x + 2)(x - 3)
f) 3x2 - 5x - 8
= 3x2 + 3x - 8x - 8
= 3x(x + 1) - 8(x + 1)
= (3x - 8)(x + 1)
g) x3 + 3x2 + 6x + 4
= (x3 + 3x2 + 3x + 1) + (3x + 3)
= (x + 1)3 + 3(x + 1)
= (x + 1)[(x + 1)2 + 3]
h) 3x3 - 5x2 - 6x + 8
= 3x3 - 3x2 - 2x2 - 6x + 8
= 3x3 - 3x2 - 2x2 + 2x - 8x + 8
= 3x2(x - 1) - 2x(x - 1) - 8(x - 1)
= (3x2 - 2x - 8)(x - 1)
a) \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(5x^2-10xy+5y^2-20z^2\) (đã sửa đề)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
c) \(x^2y-xy^2+x^3-y^3\)
\(=xy\left(x-y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\)
\(=\left(x+2\right)^2-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x-y+2\right)\)
e) \(x^2-x-6=\left(x+2\right)\left(x-3\right)\)
f) \(3x^2-5x-8\)
\(=\left(3x^2+3x\right)-\left(8x+8\right)\)
\(=3x\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-8\right)\)
Câu a trước đi ạ ^^
a) 7x - 6x2 - 2
= - 6x2 + 7x - 2
= (- 6x2 + 3x) + (4x - 2)
= 3x (- 2x + 1) + 2 (2x-1)
= - 3x ( 2x -1) + 2 (2x - 1)
= ( 2x -1 ) ( - 3x +2 )
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a.16x-5x2-3 = - ( 5x2-16x+3) = -( 5x2-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3)
b.x^3-x+3x^2y+3xy^2+y^3-y = \(\left(x^3+3x^2y+3xy^2+y^3\right)-\)\(\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\)\(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
d.x^2+x-6 = \(x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-2\right)\)
e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
f.2(x^5)-x^2-5x ( mik ko bik làm)
g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)
\(=\left(x^2-2\right)-x^2=\left(x^2-2+x\right)\left(x^2-2-x\right)\)