Ta có : x2 - 9 = 2(x + 3)2
=> x2 - 9 - 2(x + 3)2 = 0
=> x2 - 9 - 2(x2 + 6x + 9) = 0
=> x2 - 9 - 2x2 - 12x - 9 = 0
=> -x2 - 12x - 18 = 0
=> sai đề trầm trọng
\(x^2-9=2\left(x+3\right)^2\)
\(x^2-9=2\left(x^2+6x+9\right)\)
\(x^2-9=2x^2+12x+18\)
\(x^2-9-2x^2-12x-18=0\)
\(-x^2-12x-27=0\)
\(-\left(x^2+12x+27\right)=0\)
\(-\left(x^2+12x+36-9\right)=0\)
\(-\left(x^2+12x+36\right)-9=0\)
\(-\left(x+6\right)^2-3^2=0\)
\(\left(x-6\right)^2-3^2=0\)
\(\left(x-6-3\right)\left(x-6+3\right)=0\)
\(\left(x-9\right)\left(x-3\right)=0\)
\(\orbr{\begin{cases}x-9=0\\x-3=0\end{cases}}=>\orbr{\begin{cases}x=9\\x=3\end{cases}}\)
vậy \(x=9\) hoặc \(x=3\)
\(4x^2-4x+1=\left(5-x\right)^2\)
\(\left(2x-1\right)^2=\left(5-x\right)^2\)
\(2x-1=5-x\)
\(2x+x=5+1\)
\(3x=6\)
\(x=2\)
vậy \(x=2\)
