Tìm x, biết: (a) x^2−4x+4=0 . (b) x(3+x)−12=x^2.
Giải thích
a) \({x^2} - 4x + 4 = 0\)
\({\left( {x - 2} \right)^2} = 0\)
\[x - 2 = 0\]
\[x = 2\]
b) \(x\left( {3 + x} \right) - 12 = {x^2}\)
\(\begin{array}{l}3x + {x^2} - 12 = {x^2}\\3x = 12\\x = 4\end{array}\)