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`(2 1/3 + 3 1/2): (-4 1/6 + 3 1/7) +7,5`
`=(7/3 +7/2) : (-25/6 + 22/7) + 15/2`
`=35/6 : (-43/42) + 15/2`
`=-245/43+15/2`
`=155/86`
1: \(=\dfrac{-8}{11}\left(\dfrac{3}{2}+\dfrac{33}{20}+\dfrac{11}{10}\right)\)
\(=\dfrac{-8}{11}\cdot\dfrac{30+33+22}{20}=\dfrac{-8}{11}\cdot\dfrac{85}{20}=-\dfrac{34}{11}\)
2: \(=\dfrac{2}{3}+\dfrac{1}{3}=1\)
a, \(\dfrac{3}{4}x=-\dfrac{9}{8}\)
x= \(-\dfrac{3}{2}\)
b, |x| + 0,25= 5,25
|x | = 5
=> x\(\in\){ +- 5}
Ko chắc đúng, kiểm tra trc khi làm
\(\dfrac{2x-1}{3}=\dfrac{-5}{0.6}\)
\(\Leftrightarrow2x-1=-25\)
hay x=-12
1)
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)
a: \(=\dfrac{9}{13}\cdot\dfrac{4}{5}=\dfrac{36}{65}\)
b: \(=\dfrac{-7}{10}:\dfrac{3}{2}=\dfrac{-7}{10}\cdot\dfrac{2}{3}=\dfrac{-14}{30}=-\dfrac{7}{15}\)
c: \(=\dfrac{7}{6}\left(3+\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{7}{6}\cdot3=\dfrac{7}{2}\)
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
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
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)
\(M=0\times\dfrac{2022}{2021}\)
M=0
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}+\dfrac{6}{7}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}+0,875+0,7}+\dfrac{6}{7}.\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}.\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1.\)
Vậy.....
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
\(\dfrac{\dfrac{2}{3}+0,25-0,6}{\dfrac{2}{3}-0,25+0,6}:\dfrac{\dfrac{2}{5}-\dfrac{1}{6}+\dfrac{3}{7}}{\dfrac{2}{5}+\dfrac{1}{6}-\dfrac{3}{7}}\)
\(=\dfrac{\dfrac{19}{60}}{\dfrac{61}{60}}:\dfrac{\dfrac{139}{210}}{\dfrac{29}{210}}=\dfrac{19}{61}:\dfrac{139}{29}=\dfrac{551}{8479}\)
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