Question

phân tích đa thức thành nhân tử:
a) x² - 25;    b) x⁶ - y⁴;    c) 9x² - 1/16y²;     d) 81 - (3x + 2) ²

Answer 1

\(a.x^2-25=x^2-5^2=\left(x+5\right)\left(x-5\right).\)

\(b.x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3+y^2\right)\left(x^3-y^2\right).\)

\(c.9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right)\)

\(=\left(3x+\dfrac{1}{4}y\right)\left(3x-\dfrac{1}{4}y\right).\)

\(d.81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(11+3x\right)\left(-3x+7\right).\)