a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)

\(\Rightarrow2,8x-32=-60\)

\(\Rightarrow2,8x=-28\)

\(\Rightarrow x=-10\)

Vậy x = -10

b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)

\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)

\(\Rightarrow4,5-2x=\dfrac{121}{98}\)

\(\Rightarrow2x=\dfrac{160}{49}\)

\(\Rightarrow x=\dfrac{80}{49}\)

Vậy \(x=\dfrac{80}{49}\)

\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)

\(2,8x-32=-90.\dfrac{2}{3}\)

\(2,8x-32=-60\)

\(2,8x=-60+32\)

\(2,8x=-28\)

\(x=-28:2,8\)

\(x=-10\)

Vậy \(x=-10\)

\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)

\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)

\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)

\(4,5-2x=\dfrac{121}{98}\)

\(2x=4,5-\dfrac{121}{98}\)

\(2x=\dfrac{160}{49}\)

\(x=\dfrac{160}{49}:2\)

\(x=\dfrac{80}{49}\)

Vậy \(x=\dfrac{80}{49}\)

\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)

\(2,8x-32\) \(=-90.\dfrac{2}{3}\)

\(2,8x-32\)

a)-12.(8)

b) -0.6

a) \(-0,6x-\dfrac{7}{3}=5,4\Leftrightarrow-\dfrac{3}{5}x=5,4+\dfrac{7}{3}\Leftrightarrow x=\dfrac{116}{15}.\left(-\dfrac{5}{3}\right)=-\dfrac{116}{9}\).

b) \(2,8:\left(\dfrac{1}{5}-3x\right)=1\dfrac{2}{5}\Leftrightarrow\dfrac{1}{5}-3x=2,8:\dfrac{7}{5}\Leftrightarrow-3x=2-\dfrac{1}{5}\Leftrightarrow x=\dfrac{9}{5}:\left(-3\right)=-\dfrac{3}{5}\).

d: =>-x-5/6=7/12-4/12=3/12=1/4

=>-x=1/4+5/6=13/12

hay x=-13/12

e: =>x+3=-5

hay x=-8

f: =>4,5-2x=-1/2

=>2x=5

hay x=5/2

a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)

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

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

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

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

=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)

b: ĐKXĐ: x>=0

\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)

=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

a/  => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)

=> \(\dfrac{1}{x}=\dfrac{2}{5}\)

=> x = 5/2

b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)

=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)

=> \(x=\dfrac{2}{5}\)

c/ => | x + 1| = 10/21

=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)

d/ => \(5x+5=6x-3\)

=> x = 8

`a)4/5+5 1/2 xx (4,5-2)+7/10`

`=4/5+11/2*2,5+7/10`

`=0,8+2,2+0,7`

`=3+0,7=3,7`

`b)125%xx 17/4:(1 5/16-0,5)+2008`

`=1,25xx4,25:13/16+2008`

`=85/13+2008`

`=2014 7/13`

`c)5/11+(16/11+1)`

`=5/11+1+5/11+1`

`=2+10/11=32/11`

`d)3/17+11/4+5/8+14/17+3/8`

`=3/17+14/17+5/8+3/8+11/4`

`=1+1+11/4`

`=19/4`

a) 

\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)

<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)

<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)

b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)

= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)

c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)

= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)

d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)

= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)

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

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

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

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

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

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

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

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

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

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

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

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

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 =....

tham khảo

cái này đúng không nhỉ 

\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\dfrac{11}{7}\)

\(\Leftrightarrow4,5-2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=4,5-\dfrac{1}{2}\)

\(\Leftrightarrow x=4:2\)

\(\Leftrightarrow x=2\)

4,5 -2x = 11/14 : 1 4/7

4,5 -2x = 11/14 : 11/7

4,5 -2x =  1/2

2x = 4,5 - 1/2

2x = 4

x = 4 : 2

x = 2