Question

Tìm số nguyên x

a) x³ - 25x=0

b) (x²-81).(18-3x)=0

c)|2x-5|=|x+3|

Answer 1

\(x^3-25x=0\)

\(x\left(x^2-25\right)=0\)

\(x\left(x-5\right)\left(x+5\right)=0\)

\(x=0,x=5,x=-5\)

Answer 2

\(a,x^3-25x=0\)

\(\Leftrightarrow x\left[x^2-25\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy : \(x\in\left\{0;\pm5\right\}\)

Answer 3

\(b,\left[x^2-81\right]\left[18-3x\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-81=0\\18-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=81\\3x=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm9\\x=6\end{cases}}\)

Answer 4

\(c,\left|2x-5\right|=\left|x+3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=x+3\\2x-5=-x+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5-x=3\\2x-5-(-x)=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x-5=3\\2x-(-x)-5=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=3\\3x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{2}{3}\end{cases}}\)