Question

phân tích đa thức thành nhân tử

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

b)\(2xy-x^{^{ }2}-y^2+16\)

c)\(x^2-2x-4y^2-4y\)

d)\(x^2+6x+9-y^2\)

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

f)\(9x-x^{^{ }3}\)

g)\(\left(2xy+1\right)^2-\left(2x+y\right)^2\)

Answer 1

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

\(2xy-x^2-y^2+16\\ =\left(x^2-2xy+y^2\right)-16\\ =\left(x-y\right)^2-16\\ =\left(x-y+4\right)\cdot\left(x-y-4\right)\)

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

\(x^2+6x+9-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\cdot\left(x-3+y\right)\)

\(3x^2+6xy+3y^2-3z^2\\ =3\cdot\left(x^2+2xy+y^2-z^2\right)\\ =3\cdot\left[\left(x^2+2xy+y^2\right)-y^2\right]\\ =3\cdot\left[\left(x-y\right)^2-z^2\right]\\ =3\cdot\left(x-y-z\right)\cdot\left(x-y+z\right)\)

\(9x-x^3\\ =x\cdot\left(9-x^2\right)\\ =x\cdot\left(3-x\right)\cdot\left(3+x\right)\)

\(\left(2xy+1\right)^2-\left(2x+y\right)^2\\ =\left(2xy+1-2x-y\right)\cdot\left(2xy+1+2x-y\right)\)