Question

Giups e vs

bài 1 : tính :

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

b \(\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)\dfrac{ }{ }\)

c\(\left(x+\dfrac{1}{3}\right)^3\)

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

Answer 1

a: \(\left(2x-\dfrac{1}{2}\right)^3\)

\(=\left(2x\right)^3-3\cdot4x^2\cdot\dfrac{1}{2}+3\cdot2x\cdot\dfrac{1}{4}-\dfrac{1}{8}\)

\(=8x^3-6x^2+\dfrac{3}{2}x-\dfrac{1}{8}\)

b: \(\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)=\dfrac{1}{4}x^2-y^2\)

c: \(\left(x+\dfrac{1}{3}\right)^3\)

\(=x^3+3\cdot x^2\cdot\dfrac{1}{3}+3\cdot x\cdot\dfrac{1}{9}+\dfrac{1}{27}\)

\(=x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)

d: \(\left(x-2\right)\left(x^2+2x+4\right)=x^3-8\)