Tìm x, biết: (a) x^2+9x=0;

a) \({x^2} + 9x = 0\)

\(x(x + 9) = 0\)

Vậy \(x \in \left\{ {0; - 9} \right\}\)

b) \[x(2x + 4) + 2(3 - {x^2}) = 10\]

\(\begin{array}{l}2{x^2} + 4x + 6 - 2{x^2} = 10\\4x = 10 - 6\end{array}\)

\(\begin{array}{l}4x = 4\\x = 1\end{array}\)

Vậy x = 1

c) \((8{x^4} - 6{x^2}):2{x^2} - 22 = 0\)

\(\begin{array}{l}4{x^2} - 3 - 22 = 0\\4{x^2} - 25 = 0\end{array}\)

\(\begin{array}{l}4{x^2} = 25\\x = \pm \frac{5}{2}\end{array}\)

Vậy \(x \in \left\{ { - \frac{5}{2};\frac{5}{2}} \right\}\)