\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)

\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)

\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)

\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)

a) (31 6/13 + 5 9/41) - 36 6/13

= 409/13 + 214/41 - 474/13

= (409/13 - 474/13) + 214/41

= -5 + 214/41

= 9/41

b) 5/3 + (-2/7) - (-1,2)

= 5/3 - 2/7 + 6/5

= 29/21 + 6/5

= 271/105

c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)

= 1/4 + 3/5 - 1/8 + 2/5 - 5/4

= (1/4 - 5/4) + (3/5 + 2/5) - 1/8

= -1 + 1 - 1/8

= -1/8

Tính 1 câu thoy nhé !

\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)

= \(\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)

=\(\dfrac{3}{7}.-14=-6\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

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

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

a:=>3x=15

=>x=5

b: =>x+3=0,96

=>x=-2,04

c: =>x^2=36

=>x=6 hoặc x=-6

`a, 3/4=(3x)/20`

`3x*4=3*20`

`3x*4=60`

`3x=60 \div 4`

`3x=15`

`x=15 \div 3`

`x=5`

`b, (1,2)/(x+3)=5/4`

`1,2*4=(x+3)*5`

`4,8=(x+3)*5`

`x+3= 4,8 \div 5`

`x+3=0,96`

`x=0,96-3`

`x=-2,04`

`c, (x^2)/32=9/8`

`x^2*8=32*9`

`x^2*8=288`

`x^2=288 \div 8`

`x^2=36`

`x^2=(+-6)^2`

`-> \text {x= 6 hoặc -6}`

B=-9/2

C=7/2

\(B=\dfrac{2}{7}-\dfrac{6}{7}:3-3:\dfrac{2}{3}\)

\(B=\dfrac{2}{7}-\dfrac{2}{7}-\dfrac{9}{2}=0-\dfrac{9}{2}=-\dfrac{9}{2}\)

Vậy B \(=-\dfrac{9}{2}\)

\(C=1,2.\dfrac{25}{12}-\left(\dfrac{1}{2}+\dfrac{1}{3}\right):-\dfrac{5}{6}\)

\(C=\dfrac{6}{5}.\dfrac{25}{12}-\dfrac{5}{6}.\dfrac{-6}{5}\)

\(C=\dfrac{5}{2}-\left(-1\right)=\dfrac{7}{2}\)

Vậy \(C=\dfrac{7}{2}\)

a

=5/7 * (5+6-4/11)

= 5/7+7/11

=5/11

b

=1,2 - 9 + 2,5

=  -7,8 + 2,5 

= -5,3

a

=5/7 * (5+6-4/11)

= 5/7+7/11

=5/11

b

=1,2 - 9 + 2,5

=  -7,8 + 2,5 

= -5,3

a, 5/7 * (5/11+6/11-4/11)

= 5/7* (-3/11) = -15/77

a, \(\left|x+1,2\right|=0,5\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1,2=0,5\\x+1,2=-0,5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-0,7\\x=-1,7\end{matrix}\right.\)

Vậy ....

b, \(\left|x-\dfrac{1}{2}\right|+\dfrac{5}{6}=1\dfrac{1}{2}\)

\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=1\dfrac{1}{2}-\dfrac{5}{6}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)

Vậy .....

c, \(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-2,8\right|\)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=2,8\)

\(\left|x-\dfrac{1}{2}\right|=2,8-\dfrac{4}{5}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=2\\x-\dfrac{1}{2}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-8}{5}\end{matrix}\right.\)

Vậy ...