Question

Tìm số nguyên x:
a, x (2x + 3) = 0
b, x³ - 9x = 0
c, (x - 5)(2x - 6) = 0

Answer 1

a)Ta có:

\(x\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{-3}{2}\left(ktm\right)\end{matrix}\right.\)

b) Ta có:

\(x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)

c)Ta có:

\(\left(x-5\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)