Question

(x+4)(x^2+4x+16)

(1/3x+2y)(1/9x^2-2/3xy+4y^2)

(x-3y)(x^2+3xy+9y^2)

(x^2-1/3)(x^4+1/3x^2+1/9)

tính:(bài tổng hai lập phương,hiệu hai lập phương)

Answer 1

\(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}xy\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)

\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)

\(=x^6-\dfrac{1}{27}\)

Answer 2

Giải:

+) \(\left(x-4\right)\left(x^2+4x+16\right)\)

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

\(=x^3-4^3\)

\(=x^3-64\)

+) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

+) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

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

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

\(=x^3-27y^3\)

+) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)

\(=x^6-\dfrac{1}{27}\)

Vậy ...