\(\dfrac{1}{5}\) \(+\) \(\dfrac{4}{5}\)\(x\) \(=\) \(-3\)
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a: =>1/2x-3/4x=-5/6+7/3
=>-1/4x=14/6-5/6=3/2
=>x=-3/2*4=-6
b: =>4/5x-3/2x=1/2+6/5
=>-7/10x=17/10
=>x=-17/7
c: =>6/5x+6/20=6/5-1/3x
=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10
=>x=27/46
d: =>6x+3/2+4/5=1/2-2x
=>8x=1/2-3/2-4/5=-1-4/5=-9/5
=>x=-9/40
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)
\(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)
\(\dfrac{4}{5}\).\(x\) = 1
\(x\) = \(\dfrac{5}{4}\)
b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)
\(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )
\(x\) = - \(\dfrac{20}{29}\)
c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\).\(x\) = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\).\(x\) = - \(\dfrac{13}{15}\)
\(x\) = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)
\(x\) = - \(\dfrac{91}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
\(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{49}{30}\)
\(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)
\(\dfrac{8}{5}x\dfrac{5}{8}=\dfrac{1}{1}=1\)
\(\dfrac{6}{7}x\dfrac{4}{7}=\dfrac{24}{49}\)
\(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}x\dfrac{5}{4}=\dfrac{1}{1}=1\)
\(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}x\dfrac{5}{5}=\dfrac{1}{1}=1\)
1) \(\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{1+2}{3}=\dfrac{3}{3}=1\)
2) \(\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{24}{30}+\dfrac{25}{30}=\dfrac{24+25}{30}=\dfrac{49}{30}\)
3) \(\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{4-3}{5}=\dfrac{1}{5}\)
4) \(\dfrac{9}{8}-\dfrac{4}{2}=\dfrac{9}{8}-2=\dfrac{9}{8}-\dfrac{16}{8}=-\dfrac{7}{8}\)
5) \(\dfrac{8}{5}\times\dfrac{5}{8}=\dfrac{8\times5}{5\times8}=\dfrac{40}{40}=1\)
6) \(\dfrac{6}{7}\times\dfrac{4}{7}=\dfrac{6\times4}{7}=\dfrac{24}{7}\)
7) \(\dfrac{4}{5}:\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{5}{4}=\dfrac{4\times5}{5\times4}=\dfrac{20}{20}=1\)
8) \(\dfrac{5}{5}:\dfrac{5}{5}=\dfrac{5}{5}\times\dfrac{5}{5}=\dfrac{5\times5}{5\times5}=\dfrac{25}{25}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
\(\dfrac{4}{5}x=-3-\dfrac{1}{5}=\dfrac{-16}{5}\Leftrightarrow x=-\dfrac{16}{5}:\dfrac{4}{5}=-4\)
`1/5 + 4/5x = -3`
`4/5x = -3 - 1/5`
`4/5x = -15/5 - 1/5`
`4/5x=-16/5`
`x=-16/5:4/5`
`x=-16/5 xx 5/4`
`x=-4`