Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
\(1,2x^3+3x^2-8x+3\)
\(=2x^3-2x^2+5x^2-5x-3x+3\)
\(=2x^2\left(x-1\right)+5x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(2x^2+5x-3\right)\left(x-1\right)\)
\(=\left(2x-1\right)\left(x+3\right)\left(x-1\right)\)
\(2,x^3-5x^2+2x+8\)
\(=x^3+x^2-6x^2-6x+8x+8\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x^2-6x+8\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\)
\(3,-6x^3+x^2+5x-2\)
\(=-6x^3-6x^2+7x^2+7x-2x-2\)
\(=-6x^2\left(x+1\right)+7x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(-6x^2+7x-2\right)\left(x+1\right)\)
\(=\left(-6x^2-3x-4x-2\right)\left(x+1\right)\)
\(=\left[-3x\left(2x+1\right)-2\left(2x+1\right)\right]\left(x+1\right)\)
\(=\left(-3x-2\right)\left(2x+1\right)\left(x+1\right)\)
\(4,3x^3+19x^2+4x-12\)
\(=3x^3+18x^2+x^2+6x-2x-12\)
\(=3x^2\left(x+6\right)+x\left(x+6\right)-2\left(x+6\right)\)
\(=\left(3x^2+x-2\right)\left(x+6\right)\)
\(=\left(3x-2\right)\left(x+1\right)\left(x+6\right)\)
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
AI LÀM MÌNH SẼ K ĐÚNG + 1 THẺ KHÓA HỌC T.ANH GIAO TIẾP
\(a.2x^2+5x+2=2x^2+4x+x+2=\left(x+2\right)\left(2x+1\right)\)