Question

Phân tích đa thức sau thành nhân tử

a)\(x^2+4x-y^2+4\)

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

c)\(x^2-2xy+y^2-z^2+2zt-t^2\)

Giải giúp mik nhé !!Cảm ơn m.n

Answer 1

a) \(x^2+4x-y^2+4\)

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

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

Answer 2

c) \(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

Answer 3

a) x² + 4x - y² + 4 = (x² + 4x + 4) - y² = (x + 2)² - y = (x + 2 - y)(x + 2 + y)

b) 3x² + 6xy + 3y² - 3z² = 3(x² + 2xy + y² - z²) = 3[(x² + 2xy + y²) - z²) = 3[(x + y)² - z) = 3(x + y + z)(x + y - z)²

Answer 4

b) \(3x^2+6xy+3y^2-3z^2\)

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

\(=3\left[\left(x+y\right)^2-z^2\right]\)

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