Question

Tính

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

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

Answer 1

\(A=\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)=\dfrac{9}{16}x^2-4\)

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

Answer 2

Giải:

a) \(\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)\)

\(=\left(\dfrac{3}{4}x\right)^2-2^2\)

\(=\dfrac{9}{16}x^2-4\)

Vậy ...

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

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

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

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

Vậy ...