Question

tìm x\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

 

Answer 1

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)

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

\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)

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

\(2x=0\)

\(x=0:2\Rightarrow x=0\)

k cho mình nhé

Answer 2

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2-\left(\frac{3}{5}\right)^2=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}+\frac{3}{5}\right)\left(2x+\frac{3}{5}-\frac{3}{5}\right)=0\)

\(\Leftrightarrow2x\left(2x+\frac{6}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\2x+\frac{6}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}}\)

Answer 3

( 2x + 3/5 )^2 = 9/25

=> (2x + 3/5)^2 = (3/5)^2

=> 2x + 3/5 = 3/5  hoặc 2x + 3/5 = -3/5

giải từng trường hợp tìm được x = 0 hoặc x = -3/5