Question

[Dung Vu]

Phân tích đa thức thành nhân tử :

a. \(\dfrac{1}{2}x^2-2y^2\)

b. \(\dfrac{1}{3}xy+x^2z+xz\)

c. \(18x^3-\dfrac{8}{25}x\)

d. \(\dfrac{2}{5}x^2+5x^3+x^2y\)

e. \(\dfrac{1}{2}\left(x^2+y^2\right)^2-2x^2y^2\)

f. \(27x^3-\dfrac{1}{8}y^3\)

g. \(\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{32}\)

Answer 1

[Nguyễn Hoàng Minh]

\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)

\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)