=1/8x(3/2-1/4)+3/8x5/4

=1/8x5/4+3/8x5/4

=5/4x(1/8+3/8)

=5/8

\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)

\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)

\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)

\(C=\dfrac{5}{8}\)

\(\dfrac{3}{2}x-0,2=\dfrac{3}{5}\)

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

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

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

   \(x=\dfrac{4}{5}:\dfrac{3}{2}\)

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

   \(x=\dfrac{8}{15}\)

\(\dfrac{1}{3}+x=\dfrac{3}{4}\)

        \(x=\dfrac{3}{4}-\dfrac{1}{3}\)

        \(x=\dfrac{9}{12}-\dfrac{4}{12}\)

        \(x=\dfrac{5}{12}\)

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

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

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

\(\dfrac{3}{2}x=\dfrac{13}{20}\)

   \(x=\dfrac{13}{20}:\dfrac{3}{2}\)

   \(x=\dfrac{13}{20}\cdot\dfrac{2}{3}\)

   \(x=\dfrac{13}{30}\)

\(\dfrac{11}{8}-\dfrac{3}{8}\cdot x=\dfrac{1}{8}\)

          \(\dfrac{3}{8}\cdot x=\dfrac{11}{8}-\dfrac{1}{8}\)

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

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

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

                \(x=\dfrac{10}{3}\)

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

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

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

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

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}-\dfrac{7}{8}\)

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

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

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)

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

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

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

Câu D ko bt

\(\begin{array}{l}(\dfrac{1}{2}{x^4} - \dfrac{1}{4}{x^3} + x):( - \dfrac{1}{8}x) = \dfrac{1}{2}{x^4}:( - \dfrac{1}{8}x) - \dfrac{1}{4}{x^3}:( - \dfrac{1}{8}x) + x:( - \dfrac{1}{8}x)\\ = (\dfrac{1}{2}: - \dfrac{1}{8}).({x^4}:x) - (\dfrac{1}{4}: - \dfrac{1}{8}).({x^3}:x) + (1: - \dfrac{1}{8}).(x:x)\\ =  - 4.{x^{4 - 1}} - ( - 2).{x^{3 - 1}} + ( - 8).{x^{1 - 1}}\\ =  - 4{x^3} + 2{x^2} - 8\end{array}\)

giúp mình đi ạ 

\(=\dfrac{1}{x+1}-\dfrac{8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-4-8}{\left(x+1\right)\left(x-4\right)}=\dfrac{x-12}{\left(x+1\right)\left(x-4\right)}=\dfrac{-11}{2\cdot\left(-3\right)}=\dfrac{11}{6}\)

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)

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

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

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

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

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

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

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

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

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

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

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

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

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 =....