Đáp án:
\(f,\ 1000-x^3=(10-x)(100+10x+x^2)\\ j,\ \dfrac{1}{125}y^3+x^3=\bigg(\dfrac{1}{5}y+x\bigg)\bigg(\dfrac{1}{25}y^2-\dfrac{1}{5}xy+x^2\bigg)\\ k,\ \dfrac{1}{27}x^3-27y^3=\bigg(\dfrac{1}{3}x-3y\bigg)\bigg(\dfrac{1}{9}x^2+xy+9y^2\bigg)\)
Giải thích các bước giải:
\(f,\ 1000-x^3\\ =10^3-x^3\\ =(10-x)(100+10x+x^2)\\ j,\ \dfrac{1}{125}y^3+x^3\\ =\bigg(\dfrac{1}{5}y\bigg)^3+x^3\\ =\bigg(\dfrac{1}{5}y+x\bigg)\bigg(\dfrac{1}{25}y^2-\dfrac{1}{5}xy+x^2\bigg)\\ k,\ \dfrac{1}{27}x^3-27y^3\\ =\bigg(\dfrac{1}{3}x\bigg)^3-(3y)^3\\ =\bigg(\dfrac{1}{3}x-3y\bigg)\bigg(\dfrac{1}{9}x^2+xy+9y^2\bigg)\)
chúc em học tốt!
