a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)
b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)
b, 3\(x^2\) - 12 = 2\(x^2\) + 4
3\(x^2\) - 2\(x^2\) = 12 + 4
\(x^2\) = 16
\(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
c) \(2x-3\left(2x+1\right)=4x-5\left(x-3\right)\\ 2x-6x-3=4x-5x+15\\ -4x-3=-x+15\\ 4x-x=-15-3\\ 3x=-18\\ x=-18:3=-6\)
d) \(\left(x-2\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
c, 2\(x\) - 3.(2\(x\) + 1) = 4\(x\) - 5.(\(x-3\))
2\(x\) - 6\(x\) - 3 = 4\(x\) - 5\(x\) + 15
-4\(x\) - 3 = - \(x\) + 15
4\(x\) - \(x\) = - 3 - 15
3\(x\) = -18
\(x\) = - 6