\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Rightarrow4x^2+20x+25-x^2-4x-4=0\)
\(\Rightarrow3x^2+16x+21=0\)
\(\Rightarrow3x^2+9x+7x+21=0\)
\(\Rightarrow3x\left(x+3\right)+7\left(x+3\right)=0\)
=> (x + 3).(3x + 7) = 0
=> x = -3 hoặc x = -7/3
\(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Rightarrow2x+5=x+2\)
\(2x-x=-5+2\)
\(x=-3\)
Ta có: \(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{-7}{3}\end{matrix}\right.\)
