Question

Bài 1 : Phân tích đa thức thành nhân tử 
a) 9x²y² + 15x²y - 21xy²
b) x(y-z) - xy² +xz²
c) 27x³ - 1/8 y⁶ ( y⁶ ở ngoài 1/8 nhá )
d) x³ - 2x² + x
e) x² - 7x + 12
f) x⁸ + 64 
 

Answer 1

a, \(=3xy\left(3xy+5x-7y\right)\)

b, \(=x\left(y-z\right)-x\left(y^2-z^2\right)=x\left(y-z\right)-x\left(y-z\right)\left(z+y\right)=\left(y-z\right)\left(x-xz-xy\right)\)

c, \(=\left(3x\right)^3-\left(\dfrac{1}{2}y^2\right)^3=\left(3x-\dfrac{1}{2}y^2\right)\left(9x^2-\dfrac{3}{2}xy^3+\dfrac{1}{4}y^4\right)\)

d, \(=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)

e, \(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)

f, \(x^8+64=\left(x^2+4\right)\left(x^4-8x^2+16\right)\)