\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

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

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

                   \(x=\dfrac{-2}{3}\)

vậy x =....

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}=\dfrac{5}{12}.\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)=\dfrac{5}{12}.\left(-1\right)=-\dfrac{5}{12}\)

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

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

\(a.\)

\(x-\dfrac{5}{6}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}+\dfrac{5}{6}=\dfrac{3+5}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

\(b.\)

\(-\dfrac{3}{4}-x=-\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{3}{4}--\dfrac{7}{12}=-\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{\left(-3\right)\cdot3+7}{12}=\dfrac{-2}{12}=-\dfrac{1}{6}\)

a)x=1/2+5/6

x=/3/6+5/6

x=8/6=4/3

b)x=-3/4- (-7/12)

x=-9/12-(-7/12)

x=-1/6

a) \(x-\dfrac{5}{6}=\dfrac{1}{2}\) 

           \(x=\dfrac{1}{2}+\dfrac{5}{6}\) 

          \(x=\dfrac{4}{3}\) 

b) \(\dfrac{-3}{4}-x=\dfrac{-7}{12}\) 

               \(x=\dfrac{-3}{4}-\dfrac{-7}{12}\) 

               \(x=\dfrac{-1}{6}\)

a.

31,5 – x = (18,6 – 12,3) : 3

31,5 - x = 2,1

x = 31,5 - 2,1

x = 29,4

b.

???

Bài 2. Tìm x biết:

a) 31,5 – x = (18,6 – 12,3) : 3

31,5 – x = 6,3 : 3

31,5 – x = 2,1

x = 31,5 – 2,1

x = 29,4

b) Đề sai.

a, X = 29,4

a, \(x\) : \(\dfrac{13}{3}\) = -2,5

    \(x\)         = -2,5 . \(\dfrac{13}{3}\)

    \(x\)         = \(\dfrac{65}{6}\)

b,\(\dfrac{3}{5}\)\(x\)         = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)

   \(\dfrac{3}{5}x\)         = \(\dfrac{-3}{20}\)

      \(x\)          = \(\dfrac{-3}{20}\) :  \(\dfrac{3}{5}\)

      \(x\)          = \(\dfrac{-1}{4}\)

c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)

              \(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)

               \(\dfrac{12}{13}x=2\)

                    \(x=2:\dfrac{12}{13}\)

                    \(x=\dfrac{13}{6}\)

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)

a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)

b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)

c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)

d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)

\(a,\)\(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(x=\dfrac{13}{7}.\dfrac{6}{13}\)

\(x=\dfrac{6}{7}\)

b,\(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)

\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)

\(\dfrac{4}{7}.x=\dfrac{13}{15}\)

\(x=\dfrac{13}{15}:\dfrac{4}{7}\)

\(x=\dfrac{13}{15}.\dfrac{7}{4}\)

\(x=\dfrac{91}{60}\)

a: Ta có: \(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(\Leftrightarrow x=\dfrac{13}{7}\cdot\dfrac{6}{13}\)

hay \(x=\dfrac{6}{7}\)

b: Ta có: \(\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{13}{15}\)

hay \(x=\dfrac{91}{60}\)