a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)
b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)
c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)
d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)
a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)
c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)
d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)
e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)
\(\Leftrightarrow-8x+20-2=18x-45\)
\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)
f. \(\left(3x-1\right)\left(\dfrac{-1}{2x+5}\right)=0\)
\(\left[{}\begin{matrix}3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\\\dfrac{-1}{2x+5}=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=\dfrac{-5}{2}\end{matrix}\right.\)
g. \(\dfrac{6x}{12x}-\dfrac{8}{12x}=\dfrac{7x}{12x}\Leftrightarrow-x=8\Leftrightarrow x=-8\)
h. \(\dfrac{45}{60x}+\dfrac{12x}{6x}=\dfrac{10x}{60x}\Leftrightarrow2x=-45\Leftrightarrow x=\dfrac{-45}{2}\)
i) \(\dfrac{17}{12}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}-\dfrac{17}{12}\)
\(-\left|2x-\dfrac{3}{4}\right|=-\dfrac{19}{6}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{19}{6}\)
\(\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{19}{6}\\2x-\dfrac{3}{4}=-\dfrac{19}{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{47}{24}\\x=-\dfrac{29}{24}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{47}{24};x_2=-\dfrac{29}{24}\)
