\(\left(x-2,5\right)^2=\frac{4}{9}\)
\(\left(x-2,5\right)^2=\left(\frac{2}{3}\right)^2\)
\(x-2,5=\frac{2}{3}\)
\(x=\frac{2}{3}+2,5\)
\(x=\frac{19}{6}\)
\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(2x+\frac{1}{3}=\frac{-2}{3}\)
\(2x=\frac{-2}{3}-\frac{1}{3}\)
\(2x=-1\)
\(x=\frac{-1}{2}\)
\(\left(x-2,5\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-2,5=\frac{2}{3}\\x-2,5=-\frac{2}{3}\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}+\frac{5}{2}\\x=-\frac{2}{3}+\frac{5}{2}\end{cases}}\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=\frac{4}{6}+\frac{15}{6}=\frac{19}{6}\\x=-\frac{4}{6}+\frac{15}{6}=\frac{11}{6}\end{cases}}\)
Ta có:
a)\(\left(x-2,5\right)^2=\frac{4}{9}\)
\(x-2,5=\sqrt{\frac{4}{9}}\)
\(x-2,5=\frac{2}{3}\)
\(x=\frac{2}{3}+2,5=\frac{19}{6}\)
b) \(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(2x+\frac{1}{3}=\sqrt[3]{\frac{-8}{27}}\)
\(2x+\frac{1}{3}=\frac{-2}{3}\)
\(x=\left(\frac{-2}{3}-\frac{1}{3}\right)\div2=\frac{-1}{2}\)
