b, \(4x^2-25=0\)
\(\Leftrightarrow4x^2=25\)
\(\Leftrightarrow x^2=\frac{25}{4}=\left(\pm\frac{5}{2}\right)^2\)
\(\Leftrightarrow x=\pm\frac{5}{2}\)
Vậy \(x\in\left\{\frac{5}{2};-\frac{5}{2}\right\}\)
c) x3 - 4x2 + 4x = 0
=> x3 - 2x2 - 2x2 + 4x = 0
=> x2.(x - 2) - 2x.(x - 2) = 0
=> (x - 2).(x2 - 2x) = 0
=> (x - 2).x.(x - 2) = 0
=> (x - 2)2.x = 0
=> (x - 2)2 = 0 hoặc x = 0
=> x - 2 = 0 hoặc x = 0
=> x = 2 hoặc x = 0