Question

Tìm các số nguyên x :

a) x.(x + 7) = 0 d) x. (2 + x).(7 - x) = 0

b) (x + 12).(x - 3) = 0 e) (x - 1).(x + 2).(x - 3) = 0

c) (-x + 5).(3 - x) = 0 g) (x - 5) (x² - 81) = 0

h) x³ + 27 = 0

Answer 1

a) Ta có: \(x\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

Vậy: x∈{0;-7}

b) Ta có: \(\left(x+12\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)

Vậy: x∈{-12;3}

c) Ta có: \(\left(-x+5\right)\left(3-x\right)=0\)

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

Vậy: x∈{3;5}

d) Ta có: \(x\left(2+x\right)\left(7-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)

Vậy: x∈{-2;0;7}

e) Ta có: \(\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=3\end{matrix}\right.\)

Vậy: x∈{-2;1;3}

g) Ta có: \(\left(x-5\right)\left(x^2-81\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x^2-81=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x^2=81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\\x=-9\end{matrix}\right.\)

Vậy: x∈{-9;5;9}

h) Ta có: \(x^3+27=0\)

\(\Leftrightarrow x^3=-27\)

hay x=-3

Vậy: x=-3