(x-3)2 - (2x-1)(2x+1) = 10
<=> x2 - 6x + 9 - 4x2 + 1 = 10
<=> -3x2 - 6x = 0
<=> x2 + 2x = 0
<=> x(x + 2) = 0
<=> \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
vậy ....
\(4(x-3)\)\(^2\) \(-(2x-1)(2x+1)=10\)
\(=> 4(x\)\(^2\) \(-6x+9)-4x\)\(^2\)\(-1=10\)
\(=>\)\(4x\)\(^2\)\(-24x+36-4x\)\(^2\) \(+1=10\)
\(=>-24x+27=10\)
\(=>-24x=-27\)
\(=>x=\)\(\dfrac{-27}{-24}\)\(=\) \(\dfrac{9}{8}\)
\(Vậy \)\(X=\)\(\dfrac{9}{8}\)
Nhớ vote cho mik hen