b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)

<=>\(\left|\dfrac{1}{2}-2x\right|=\dfrac{5}{3}< =>\left[{}\begin{matrix}\dfrac{1}{2}-2x=\dfrac{5}{3}\\\dfrac{1}{2}-2x=\dfrac{-5}{3}\end{matrix}\right.< =>\left[{}\begin{matrix}2x=\dfrac{-7}{6}\\2x=\dfrac{13}{6}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-7}{12}\\x=\dfrac{13}{12}\end{matrix}\right.\)

\(\left|\dfrac{1}{2}+2x\right|+\dfrac{2}{3}=\dfrac{7}{3}\)

\(\left|\dfrac{1}{2}+2x\right|=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)

⇔\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=\dfrac{5}{3}\\\dfrac{1}{2}+2x=-\dfrac{5}{3}\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)

Vậy ...

A

B

\(a,\left|2x+\dfrac{1}{2}\right|=0\\ \Leftrightarrow2x+\dfrac{1}{2}=0\\ \Leftrightarrow2x=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{4}\\ b,\left|3x+\dfrac{3}{4}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{3}{4}=3\\3x+\dfrac{3}{4}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{9}{4}\\3x=-\dfrac{15}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

C

C

5/6 và 4/6

\(\dfrac{5}{6}=\dfrac{5}{6}\)

\(\dfrac{2}{3}=\dfrac{2x2}{3x2}=\dfrac{4}{6}\)

=>36/18>x>8/15

=>2>x>8/15

mà x nguyên

nên x=1

-2

-19/3

 k mình nha

a: =5/6(1/8+2/3)=5/6*19/24=95/144

b: =3/4(7/5-1/2)=27/40

c: =35/24*2/3=35/36

\(y=\dfrac{23}{15}\)

23/15