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\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
\(3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2\left(x-2y+3\right)\)
\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
\(12x^2y-18xy^2-3xy^2=3xy\left(4x-6y-y\right)\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(y\left(x-z\right)+7\left(z-x\right)=y\left(x-z\right)-7\left(x-z\right)=\left(x-z\right)\left(y-7\right)\)
\(27x^2\left(y-1\right)-9x^3\left(1-y\right)=27x^2\left(y-1\right)+9x^3\left(y-1\right)=9x^2\left(y-1\right)\left(3-x\right)\)
a. \(x^3-2x^2+x\)
\(=x^3-x^2-x^2+x\)
\(=x^2\left(x-1\right)-x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x\right)\)
\(=\left(x-1\right)x\left(x-1\right)\)
\(=x\left(x-1\right)^2\)
b. \(x^2-2x-15\)
\(=\left(x^2-2x+1\right)-16\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-1-4\right)\left(x-1+4\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
c. \(5x^2y^3-25x^3y^4+10x^3y^3\)
\(=5x^2y^3\left(1-5xy+2x\right)\)
d. \(12x^2y-18xy^2-30y^2\)
\(=6y\left(2x^2-3xy-5y\right)\)
e. \(5\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(5-y\right)\)
a) \(6x^3-12x^2y^2+6xy^3=6x.\left(x^2-2xy^2+y^3\right)\)
b) \(\left(x^2+4\right)^2-16=\left(x^2+4-4\right)\left(x^2+4+4\right)=x^2\left(x^2+8\right)\)
c) \(5x^2-5xy-10x+10y=\left(5x^2-5xy\right)-\left(10x-10y\right)=5x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-10\right)=5\left(x-y\right)\left(x-2\right)\)
d) \(a^3-3a+3b-b^3=\left(a^3-b^3\right)-\left(3a-3b\right)=\left(a-b\right)\left(a^2+ab+b^2\right)-3.\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2+ab+b^2-3\right)\)
e) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-1-y\right)\left(x-1+y\right)\)
f) \(x^2-x-2=x^2+x-2x-2=\left(x^2+x\right)-\left(2x+2\right)=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
g) \(x^4-5x^2+4=x^4-4x^2+4-x^2=\left(x^4-4x^2+4\right)-x^2=\left(x^2-2\right)^2-x^2\)
\(=\left(x^2-2-x\right)\left(x^2-2+x\right)\)
j) \(x^3-x^3-2x^2-x=-2x^2-x=-\left(2x^2+x\right)=-x\left(2x+1\right)\)
k) \(\left(a^3-27\right)-\left(3-a\right)\left(6a+9\right)=\left(a-3\right).\left(a^2+3a+9\right)+\left(a-3\right)\left(6a+9\right)\)
\(\left(a-3\right)\left(a^2+3a+9+6a+9\right)=\left(a-3\right)\left(a^2+9a+18\right)\)
h) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)
\(=\left(x^2y-y^2x\right)-\left(x^2z-y^2z\right)+\left(z^2x-z^2y\right)\)
\(=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left[\left(xy-zx\right)-\left(zy-z^2\right)\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)
\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
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
1)
a) (x+y)3-(x+y)= (x+y)(x+y-1)
b) xem lại đề câu B nha bạn
2)
a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc=0
(a+b)3+c3-3ab(a+b+c)=0
(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)=0
(a+b+c)(a2+b2+c2-xy-yz-xz)=0
Suy ra: a3+b3+c3=3abc
1. a) = (x+y)3 -(x+y) =(x+y)((x+y)2 -1)
= (x+y)(x+y+1)(x+y-1)
b) = 5(( x-y)2 - 4z2)
= 5( x-y +2z)(x-y-2z)
2. áp dụng ( a+b+c)3 = .....rồi biến đổi
\(\text{a) }x^3y^3+x^2y^2+4\)
\(=x^3y^3+2x^2y^2-x^2y^2+4\)
\(=\left(x^3y^3+2x^2y^2\right)-\left(x^2y^2-4\right)\)
\(=x^2y^2\left(xy+2\right)-\left(xy+2\right)\left(xy-2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
\( {c)}\)\(x^4+x^3+6x^2+5x+5\)
\(=\left(x^4+x^3+x^2\right)+\left(5x^2+5x+5\right)\)
\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2+5\right)\)
\({d)}\)\(x^4-2x^3-12x^2+12x+36\)
\(=\left(x^4-2x^3-6x^2\right)-\left(6x^2-12x-36\right)\)
\(=x^2\left(x^2-2x-6\right)-6\left(x^2-2x-6\right)\)
\(=\left(x^2-2x-6\right)\left(x^2-6\right)\)
Câu b sai đề thì phải ah
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)
Phân tích đa thức thành nhân tử:
a) 5x - 5y
= 5(x - y)
b) 3xy2 + x2y
= xy(3y + x)
c) 12x2y - 18xy2 - 30y3
= 2y(6x2 - 9xy - 15y2)
= 2y(6x2 + 6xy - 15xy - 15y2)
= 2y[6x(x + y) - 15y(x + y)]
= 2y(x + y)(6x - 15y)
= 6y(x + y)(2x - 5y)
d) -17x3y - 34x2y2 + 51xy3
= -17xy(x2 + 2xy - 3y2)
= -17xy(x2 - xy + 3xy - 3y2)
= -17xy[x(x - y) + 3y(x - y)]
= -17xy(x - y)(x + 3y)
e) x(y - 1) + 3(y - 1)
= (y - 1)(x + 3)
f) 162(x - y) - 10y(y - x)
= 162(x - y) + 10y(x - y)
= (x - y)(162 + 10y)
= (x - y)(256 + 10y)
a) Ta có: 5x-5y
=5(x-y)
b) Ta có: \(3xy^2+x^2y\)
\(=xy\left(3y+x\right)\)
c) Ta có: \(12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left(2x^2-5xy+2xy-5y^2\right)\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
d) Ta có: \(-17x^3y-34x^2y^2+51xy^3\)
\(=-17xy\left(x^2+2xy-3y^2\right)\)
\(=-17xy\left(x^2+3xy-xy-3y^2\right)\)
\(=-17xy\left[x\left(x+3y\right)-y\left(x+3y\right)\right]\)
\(=-17xy\left(x+3y\right)\left(x-y\right)\)
e) Ta có: x(y-1)+3(y-1)
=(y-1)(x+3)