Question

\(a)\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)

\(b)\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)

\(_{c)\left(2x-3\right)^6=\left(2x-3\right)^6}\)

Answer 1

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

\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow x=1\)

Answer 2

b) \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{3}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Rightarrow x=\frac{1}{6}\)

Answer 3

\(a.\)\(\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)

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

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{2}\)

\(\Leftrightarrow x=1\)

\(b.\)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{3}\right)^2\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{1}{6}\)