b) Áp dụng hđt `a^2-2ab+b^2=(a-b)^2` với `a=2x+3; b=2x-5`
`(2x+3)^2-2(2x+3)(2x-5)+(2x-5)^2=x^2+6x+64`
`⇒(2x+3-2x+5)^2=x^2+6x+64`
`⇒x^2+6x+64=64`
`⇒x^2+6x=0`
`⇒x(x+6)=0`
\(⇒\left[ \begin{array}{l}x=0\\x+6=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\x=-6\end{array} \right.\)
Vậy `x=0; -6`