Question

[capricon]

Mọi người giúp mik với.

Answer 1

[Lấp La Lấp Lánh]

Bài 2:

a) \(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{3}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{3}{2}\\x-\dfrac{1}{3}=-\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

b) \(\Rightarrow2x=\dfrac{\left(-4\right).9}{3}=-12\Rightarrow x=-6\)

c) \(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

d) \(\Rightarrow x-\dfrac{1}{3}=\dfrac{1}{3}\Rightarrow x=\dfrac{2}{3}\)

3) \(\dfrac{10n+7}{5n-1}=\dfrac{2\left(5n-1\right)}{5n-1}+\dfrac{9}{5n-1}=2+\dfrac{9}{5n-1}\in Z\)

\(\Rightarrow\left(5n-1\right)\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\)

Do \(n\in Z\)

\(\Rightarrow n\in\left\{0;2\right\}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

4: Để A nguyên thì \(5n-1\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Leftrightarrow5n\in\left\{0;10\right\}\)

hay \(n\in\left\{0;2\right\}\)