\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Chúc bạn học tốt !!!
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2-\left(\frac{3}{5}\right)^2=0\)
Áp dụng bất đẳng thức A2 - B2 = ( A - B )( A + B )
Ta có :
\(\left(2x+\frac{3}{5}-\frac{3}{5}\right)\cdot\left(2x+\frac{3}{5}+\frac{3}{5}\right)=0\)
\(2x\left(2x+\frac{6}{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x+\frac{6}{5}=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
Vậy .......
