Question

Answer 1

\(a,\left(y-\dfrac{2}{3}\right)^2=y^2-2.y.\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2=y^2-\dfrac{4}{3}y+\dfrac{4}{9}\)

\(b,\left(6a+b\right)^2=\left(6a\right)^2+2.6a.b+b^2=36a^2+12ab+b^2\)

\(c,\left(\dfrac{2}{3}-a\right).\left(\dfrac{2}{3}+a\right)=\left[\left(\dfrac{2}{3}\right)^2-a^2\right]=\left(\dfrac{4}{9}-a^2\right)\)

\(d,\left(b-\dfrac{3}{2}\right)^2=b^2-2.b.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2=b^2-3b+\dfrac{9}{4}\)

Hoctot

Answer 2

a) \(\left(y-\dfrac{2}{3}\right)^2=y^2-\dfrac{4}{3}y+\dfrac{4}{9}\)

b) \(\left(6a+b\right)^2=36a^2+12ab+b^2\)

c) \(\left(\dfrac{2}{3}-a\right)\left(\dfrac{2}{3}+a\right)=\dfrac{4}{9}-a^2\)

d) \(\left(b-\dfrac{3}{2}\right)^2=b^2-3b+\dfrac{9}{4}\)