Question

\((x+\frac{1}{2})\times(\frac{2}{3}-2x)=0\)

\(\left\{x+\frac{1}{5}\right\}^2+\frac{17}{25}=\frac{26}{25}\)

Answer 1

\(\left(x+\frac{1}{2}\right)\cdot\left(\frac{2}{3}-2x\right)=0\)

TH1 : \(\Rightarrow x+\frac{1}{2}=0\)

\(x=0-\frac{1}{2}\)

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

TH2 : \(\Rightarrow\frac{2}{3}-2x=0\)

\(2x=0+\frac{2}{3}=\frac{2}{3}\)

\(x=\frac{2}{3}\div2=\frac{1}{3}\)

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

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

\(x=\frac{3}{5}^2-\frac{1}{5}^2\)

\(x=\frac{2}{5}\)