phân tích đa thức sau thành nhân tử: x3+2x2y+xy2-9x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x3 + 2x2y + xy2 – 9x
(Có x là nhân tử chung)
= x(x2 + 2xy + y2 – 9)
(Có x2 + 2xy + y2 là hằng đẳng thức)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
(Xuất hiện hằng đẳng thức (3)]
= x(x + y – 3)(x + y + 3)
a) x3 + 2x2y + xy2 – 4x = x(x2 + 2xy + y2– 4) = x[(x+y)2-4]
= x(x + y + 2)(x + y – 2)
a: \(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+1\right)\)
b: \(=\left(x-4\right)\left(x+3\right)\)
e: =(x+3)(x-2)
a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)
b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)
c) \(=\left(2x\right)^2-\left(x^2+1\right)^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=4xy\left(y-3x+2\right)\)
e) \(=x\left(x-2\right)+3\left(x-2\right)=\left(x-2\right)\left(x+3\right)\)
f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)
g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)
h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)
i) \(=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)=\left(x-3\right)^2\left(x+3\right)\)
a) x² - 10x + 25
= x² - 2.x.5 + 5²
= (x - 5)²
b) (2x - 3)² - x²
= (2x - 3 - x)(2x - 3 + x)
= (x - 3)(3x - 3)
= 3(x - 3)(x - 1)
c) x³ + 2x²y+ xy² - 9x
= x(x² + 2xy + y² - 9)
= x[(x + y)² - 3²]
= x(x + y - 3)(x + y + 3)
a/\(x^2-10x+25\)
\(=\left(x-5\right)^2\)
b/\(\left(2x-3\right)^2-x^2\)
\(=\left(2x-3-x\right)\left(2x-3+x\right)\)
\(=\left(x-3\right)\left(3x-3\right)\)
\(=3\left(x-3\right)\left(x-1\right)\)
c/\(x^3+2x^2+xy^2-9x\)
\(=x\left(x^2+2x+y^2-9\right)\)
\(=x\left[\left(x+y\right)^2-3^2\right]\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
#sdboy2mai
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)
\(=x\left(x^2+2xy+y^2-9\right)\)
=x(x+y-3)(x+y+3)
\(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-3^2\right]=x\left(x+y-3\right)\left(x+y+3\right)\)