(x+$\frac{1}{5}$)²+$\frac{17}{25}$=$\frac{52}{50}$
⇔(x+$\frac{1}{5}$)²+$\frac{17}{25}$=$\frac{26}{25}$
⇔(x+$\frac{1}{5}$)²=$\frac{26}{25}$-$\frac{17}{25}$
⇔(x+$\frac{1}{5}$)²=$\frac{9}{25}$
⇒x+$\frac{1}{5}$=$\frac{3}{5}$ và x+$\frac{1}{5}$=-$\frac{3}{5}$
TH1: x+$\frac{1}{5}$=$\frac{3}{5}$ ta có:
x=$\frac{3}{5}$-$\frac{1}{5}$
x=$\frac{2}{5}$
TH2: x+$\frac{1}{5}$=-$\frac{3}{5}$ ta có:
x=-$\frac{3}{5}$-$\frac{1}{5}$
x=-$\frac{4}{5}$
Vậy x=$\frac{2}{5}$ Hoặc x=-$\frac{4}{5}$
