\(\frac{-4}{5}\)-x:\(\frac{7}{10}\)=\(\frac{3}{50}\)
ai làm được mình cho tick
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Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
\(\frac{2}{5}:x=-\frac{1}{4}\)
\(x=\frac{2}{5}:-\frac{1}{4}\)
\(x=\frac{2}{5}\cdot-\frac{4}{1}\)
\(x=-\frac{8}{5}\)
Vậy \(x=-\frac{8}{5}\)
\(\frac{4}{7}\cdot x-\frac{2}{3}=\frac{1}{5}
\)
\(\frac{4}{7}\cdot x=\frac{1}{5}+\frac{2}{3}\)
\(\frac{4}{7}\cdot x=\frac{3}{15}+\frac{10}{15}\)
\(\frac{4}{7}\cdot x=\frac{13}{15}\)
\(x=\frac{13}{15}:\frac{4}{7}\)
\(x=\frac{13}{15}\cdot\frac{7}{4}\)
\(x=\frac{91}{60}\)
Vậy \(x=\frac{91}{60}\)
2/5 : x = -1/4
=> x = 2/5 : -1/4
=> x = -8/5
4/7.x - 2/3 = 1/5
=> 4/7x = 1/5-2/3
=> 4/7x = -7/15
=> x = -49/60
a, \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{7}{6}\)
\(\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{47}{30}\)
\(\frac{3}{2}-\frac{4}{15}x=\frac{47}{50}\)
\(\frac{4}{15}x=\frac{14}{25}\)
\(x=\frac{21}{10}\)
\(\frac{5}{12}.\left(\frac{-3}{4}\right)-\frac{7}{12}.\frac{3}{4}\)
\(=\left(\frac{-5}{12}-\frac{7}{12}\right).\frac{3}{4}\)
\(=\left(-1\right).\frac{3}{4}=\frac{-3}{4}\)
~ Hok tốt ~
\(\frac{5}{12}.\left(-\frac{3}{4}\right)-\frac{7}{12}.\frac{3}{4}\)
\(=\frac{3}{4}.\left(-\frac{5}{12}-\frac{7}{12}\right)\)
\(=\frac{3}{4}.\left(-1\right)\)
\(=-\frac{3}{4}\)
Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)
\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)
\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)
\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)
= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + ( \(2\frac{5}{7}\)+ \(4\frac{5}{7}\))
= \(4\frac{4}{5}\) + \(6\frac{5}{7}\)
= \(\frac{24}{5}\) + \(\frac{47}{7}\)
= ...... ( tính nốt nhé )
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
x:7/10=-4/5-3/50
x:7/10=-43/50
x=43/50.7/10
x=-301/500
Ta có: \(\frac{-4}{5}-x:\frac{7}{10}=\frac{3}{50}\)
\(\Rightarrow\)\(x:\frac{7}{10}=\frac{-43}{50}\)
\(\Rightarrow\)\(x=\frac{-301}{500}\)
Vậy\(x=\frac{-301}{500}\)