\(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Rightarrow2x-3=x+5\)
\(\Rightarrow2x-x=3+5\)
\(\Rightarrow x=8\)
Theo bài ra , ta có :
( 2x - 3 )2 = ( x + 5 )2
=) 2x - 3 = x + 5
=) 2x - x = 5 + 3
=) x = 8
Vậy x = 8
(2x-3)2 =(x+5)2
\(\Leftrightarrow\left(2x-3+x+5\right)\left[\left(2x-3\right)-\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
