Question

Phân tích đa thức thành nhân tử

a) x²+6xy²+9y⁴

b) x^6+y²-2x³y

C) a⁴-b⁴

D) x³-3x²+3x-1

E) 64x^6-125y^6

Giải dùm e vs ạ

Answer 1

a, (x+3y²)²

b,(x³-y)²

c,(a²+b²)(a²-b²)

d,x^3 - 3x^2 + 3x - 1
=(x^3-1)-(3x^2-3x)
=(x-1)(x^2+x+1)-3x(x-1)
=(x-1)(x^2+x+1-3x)
=(x-1)(x^2-2x+1)
=(x-1)(x-1)^2
=(x-1)^3

e, 64x⁶-125y⁶

= 2⁶x⁶ - 5³y⁶

=(4x²)³-(5y²)³

=(4x²-5y²)((4x²)²+4x²5y²+(5y²)³)

= (4x² - 5y²) • (16x⁴ + 20x²y² + 25y⁴)

Answer 2

a, \(x^2+6xy+9y^4\)

= \(x^2+2.x.3y+(3y^2)^2\)

= \((x+3y)^2\)

b,\(x^6+y^2-2x^3y\)

= \((x^3)^2-2x^3y+y^2\)

= \((x^3)^2-2.x^3.y+y^2\) = \((x^3-y)^2\)

c, \(a^4-b^4\)

= \((a^2)^2-(b^2)^2 \)

= \((a^2-b^2)(a^2+b^2)\)

d,\(x^3-3x^2+3x-1\)

= \(x^3-3.x^2.1+3.x-1^3\)

= \((x-1)^3\)

e, \(64x^6-125y^6\)

= \((4x^2)^3-(5y^2)^3\)

= \((4x^2-5y^2)[(4x^2)^2+2.4x^2.5y^2+(5y^2)^2]\)

= \((4x^2-5y^2)(16y^4+40x^2y^2+25y^4)\)