Question

[Sakura]

Tìm số nguyên x biết:

a) (3x+9)(x²-16)=0

b) (x-3)(x³+27)=0

Answer 1

[Không Tên]

a)    \(\left(3x+9\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\)\(3\left(x+3\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\x^2-16=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\pm4\end{cases}}\)

Vậy...

b)   \(\left(x-3\right)\left(x^3+27\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=0\\x^3+27=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy...

Answer 2

[Sakura]

mk cảm ơn bn