a.

\(\dfrac{x}{9}=\dfrac{4}{x}\)

\(\Rightarrow x^2=4.9\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)

b.

\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=3^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

\(\dfrac{x}{9}=\dfrac{4}{x}\)

\(x^2=4.9\)

\(x^2=36\)

\(x^2=6^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

\(---------\)

\(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=3.3=3^2\)

\(\Rightarrow\left(1\right):x+1=3\)

            \(x=3-1\Rightarrow x=2.\)

\(\Rightarrow\left(2\right):x+1=-3\)

             \(x=-3-1\Rightarrow x=-4\)

Từ \(\left(1\right)\) và \(\left(2\right)\), ta suy ra:

\(\Rightarrow x\in\left\{{}\begin{matrix}2\\-4\end{matrix}\right.\)

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

\(TH1:x\ge0\)

\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)

\(TH2:x< 0\)

\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)

\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)

Vậy ...

Giải:

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(TH1:x\ge0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

  \(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)   

\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

         \(x.\dfrac{1}{6}=\dfrac{5}{6}\) 

              \(x=\dfrac{5}{6}:\dfrac{1}{6}\) 

              \(x=5\) 

\(TH2:x\le0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\) 

\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

     \(x.\dfrac{-11}{6}=\dfrac{5}{6}\) 

               \(x=\dfrac{5}{6}:\dfrac{-11}{6}\) 

               \(x=\dfrac{-5}{11}\) 

Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

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

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

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

                 \(x=\dfrac{8}{21}\)

\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)

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

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

           \(x=-\dfrac{7}{6}-\dfrac{3}{4}\)

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Quy đồng :

\(\dfrac{-20}{36}< \dfrac{x}{36}< \dfrac{-18}{36}\)

=> x = -19

\(-20< x< -18\Rightarrow x\in\left\{-19\right\}\)

\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)

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

hay x=0

\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)

\(=\dfrac{-3}{2\sqrt{x}+4}\)

Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)

hay \(0\le x< \dfrac{25}{4}\)

a) Vì \(\dfrac{x}{y} = \dfrac{5}{3} \Rightarrow \dfrac{x}{5} = \dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\begin{array}{l}\dfrac{x}{5} = \dfrac{y}{3} = \dfrac{{x + y}}{{5 + 3}} = \dfrac{{16}}{8} = 2\\ \Rightarrow x = 2.5 = 10\\y = 2.3 = 6\end{array}\)

Vậy x=10, y=6

b) Vì \(\dfrac{x}{y} = \dfrac{9}{4} \Rightarrow \dfrac{x}{9} = \dfrac{y}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

 \(\begin{array}{l}\dfrac{x}{9} = \dfrac{y}{4} = \dfrac{{x - y}}{{9 - 4}} = \dfrac{{ - 15}}{5} =  - 3\\ \Rightarrow x = ( - 3).9 =  - 27\\y = ( - 3).4 =  - 12\end{array}\)

Vậy x = -27, y = -12.

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

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

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

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

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

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

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)