Question

Phân tích đa thức thành nhân tử :

a, x² - 5x

b, x² - 10x + 25 - y²

c, x³ + y³ - 3x - 3y

d, x³ + 2x²y + xy² - 4x

Answer 1

\(a,x^2-5x=x\left(x-5\right)\)

\(b,x^2-10x+25-y^2=\left(x^2-10x+25\right)-y^2=\left(x-5\right)^2-y^2=\left(x-5-y\right).\left(x-5+y\right)\)

\(b,x^3+y^3-3x-3y=\left(x+y\right).\left(x^2+xy+y^2\right)-3.\left(x+y\right)\)

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

\(d,x^3+2x^2y+xy^2-4x=x\left(x^2+2xy+y^2-4\right)\)

\(=x.\left(x+y-2\right).\left(x+y+2\right)\)

Answer 2

a)

\(x^2-5x=x\left(x-5\right)\)

b)

\(=\left(x^2-10x+25\right)-y^2=\left(x-5\right)^2-y^2=\left(x-5-y\right)\left(x-5+y\right)\)

c)

\(=\left(x^3+y^3\right)-\left(3x+3y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2-3\right)\)

d)

\(=x\left(x^2+2xy+y^2-4\right)\)

\(=x\left[\left(x^2+2xy+y^2\right)-2^2\right]\)

\(=x\left[\left(x+y\right)-2^2\right]\)

\(=x\left(x+y-2\right)\left(x+y+2\right)\)

Answer 3

a/x²-5x=x(x-5)

b/x²-10x+25-y²=(x²-10+25)-y²=(x-5)²-y²=(x-5-y)(x-5+y)