a) \(\dfrac{3}{5}+x=\dfrac{4}{3}\) 

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

\(\Leftrightarrow x=\dfrac{20}{15}-\dfrac{9}{15}\) 

\(\Leftrightarrow x=\dfrac{11}{15}\)

b) \(x+\dfrac{5}{6}=\dfrac{1}{7}\)

\(\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}\)

\(\Leftrightarrow x=\dfrac{6}{42}-\dfrac{35}{42}\)

\(\Leftrightarrow x=-\dfrac{29}{42}\)

c) \(x\times\dfrac{3}{8}=\dfrac{3}{4}\)

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

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

\(\Leftrightarrow x=2\)

d) \(\dfrac{4}{5}\times x=\dfrac{3}{7}\)

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

\(\Leftrightarrow x=\dfrac{3}{7}\times\dfrac{5}{4}\)

\(\Leftrightarrow x=\dfrac{15}{28}\)

a, \(\dfrac{3}{5}+x=\dfrac{4}{3}\Leftrightarrow x=\dfrac{4}{3}-\dfrac{3}{5}=\dfrac{20-9}{15}=\dfrac{11}{15}\)

b, \(x+\dfrac{5}{6}=\dfrac{1}{7}\Leftrightarrow x=\dfrac{1}{7}-\dfrac{5}{6}=\dfrac{6-35}{42}=\dfrac{-29}{42}\)

c, \(\dfrac{3x}{8}=\dfrac{3}{4}\Leftrightarrow\dfrac{3x}{8}=\dfrac{6}{8}\Rightarrow3x=6\Leftrightarrow x=2\)

d, \(\dfrac{4x}{5}=\dfrac{3}{7}\Leftrightarrow\dfrac{28x}{35}=\dfrac{15}{35}\Rightarrow x=\dfrac{15}{28}\)

a, \(\dfrac{3}{5}\) + \(X\) = \(\dfrac{4}{3}\)

           \(X\) = \(\dfrac{4}{3}\) - \(\dfrac{3}{5}\)

           \(X\) = \(\dfrac{11}{15}\)

b, \(X\) + \(\dfrac{5}{6}\) = \(\dfrac{1}{7}\)

    \(X\)        = \(\dfrac{1}{7}\) - \(\dfrac{5}{6}\)

    \(X\)        = -\(\dfrac{29}{42}\)

c, \(X\) x \(\dfrac{3}{8}\) = \(\dfrac{3}{4}\)

    \(X\)       = \(\dfrac{3}{4}\) : \(\dfrac{3}{8}\)

    \(X\)       = 2

d, \(\dfrac{4}{5}\) x \(X\) = \(\dfrac{3}{7}\)

          \(X\) = \(\dfrac{3}{7}\) : \(\dfrac{4}{5}\)

          \(X\) = \(\dfrac{15}{28}\)

#Yu

\(a,\Rightarrow x^2+4x+25-x^2=3\\ \Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\\ b,\Rightarrow\left(2x-3-4x-3\right)\left(2x-3+4x+3\right)=0\\ \Rightarrow6x\left(-2x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)

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

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

    x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)  

    x =  \(\dfrac{11}{12}\) 

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

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

    x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\) 

    x = \(\dfrac{25}{6}\)

nhanh lên nhé 

sos

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

a) x = 5/8 : 3/4

x = 5/6

b) x = 5/6 - 1/12

x = 3/4

a)x=\(\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{6}\)

b) x=\(\dfrac{5}{6}-\dfrac{1}{12}=\dfrac{3}{4}\)

a \(x=\dfrac{5}{8}:\dfrac{3}{4}\)

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

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

b \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

\(x=\dfrac{10}{12}-\dfrac{1}{12}\)

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

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

Giấy hơi nhàu, thông cảm :<

`#040911`

`a)`

`6 \times x - 5 = 613`

`=> 6 \times x = 613 + 5`

`=> 6 \times x = 618`

`=> x = 618 \div 6`

`=> x = 103`

Vậy, `x = 103`

`b)`

`12 \times x + 3 \times x = 30`

`=> x \times (12 + 3) = 30`

`=> x \times 15 = 30`

`=> x = 30 \div 15`

`=> x = 2`

Vậy, `x = 2`

`c)`

`125 - 25 \times (x - 1) = 100`

`=> 25 \times (x - 1) = 125 - 100`

`=> 25 \times (x - 1) = 25`

`=> x - 1 = 25 \div 25`

`=> x - 1 = 1`

`=> x = 1 + 1`

`=> x = 2`

Vậy, `x = 2`

`d)`

`(x - 2) \times (9x - 4) = 0?`

`=>`

TH1: `x - 2 = 0`

`=> x = 0 + 2`

`=> x = 2`

TH2: `9x - 4 = 0`

`=> 9x = 4`

`=> x = 4/9`

Vậy, `x \in {2; 4/9}.`

\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

a) \(x+\dfrac{4}{15}=\dfrac{4}{12}\)

\(x=\dfrac{4}{12}-\dfrac{4}{15}\)

\(x=\dfrac{20}{60}-\dfrac{16}{60}\)

\(x=\dfrac{1}{15}\)

b) \(x-\dfrac{5}{8}=1\dfrac{2}{3}\)

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

\(x=\dfrac{5}{3}+\dfrac{5}{8}\)

\(x=\dfrac{40}{24}+\dfrac{15}{24}\)

\(x=\dfrac{55}{24}\)