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 \(x=\dfrac{91}{60}\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

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

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)

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

                 \(\dfrac{4}{5}\).\(x\) = 1

                      \(x\) = \(\dfrac{5}{4}\)

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

              \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)

                \(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)

                  \(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )

                  \(x\) = - \(\dfrac{20}{29}\)

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

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

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

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

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

d, \(\dfrac{5}{7}\): \(x\) - 1 = \(\dfrac{2}{3}\)

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

       \(\dfrac{5}{7}\): \(x\)   = \(\dfrac{5}{3}\)

              \(x\)  = \(\dfrac{5}{7}\): \(\dfrac{5}{3}\)

               \(x\) = \(\dfrac{3}{7}\)

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

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

               \(\dfrac{12}{13}x=2\)

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

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

\(a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\) 

\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\) \(=\dfrac{1}{35}+\dfrac{1}{35}=\dfrac{2}{35}\) 

\(b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}\)

a, 2/5 + 3/4 : x = -1/2

3/4 : x = -1/2 - 2/5

3/4 : x = -9/10

x = 3/4 : -9/10

x = -5/6

b, 5/7 - 2/3 . x = 4/5 

2/3 . x = 4/5 + 5/7

2/3 . x = 53/35

x = 53/35 : 2/3

x = 159/70

c và d mình làm dược nhưng ko ghi được cái suy ra

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)

Hơi choáng

Bạn đăng 2 câu 1 lần đc ko

ai giúp mik ik T_T

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)