Question

Tìm số nguyên x biết :

a) \(-1< \dfrac{5x}{13}< 0\)

b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

Answer 1

a) \(-1< \dfrac{5x}{13}< 0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{5x}{13}>-1\\\dfrac{5x}{13}< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>-\dfrac{13}{5}\\x< 0\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{13}{5},0\right\}\)

b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

\(\Rightarrow3\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)

\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)

\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Leftrightarrow3x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{18}\)

Vậy \(x=\dfrac{1}{18}\)

Answer 2

Xử câu b trc =)

b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)

\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3=-\dfrac{1}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\)

\(\Leftrightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)

Answer 3

a)\(\dfrac{-13}{13}< \dfrac{5x}{13}< \dfrac{0}{13}\)

=)-13<5x<0

=)5x=-10;-5

x=-1;-2