2.4:(-1/2-x)=hỗn số 1 và 3/5
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17 tháng 4 2019
2
\(S1=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{100.102}\)
\(S1=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\left(\frac{51}{102}-\frac{1}{102}\right)\)
\(S1=\frac{1}{2}.\frac{25}{51}\)
\(S1=\frac{25}{102}\)
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14 tháng 7 2020
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Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
=1,4 x\(\dfrac{15}{49}-\) \(\left(\dfrac{4}{5}+\dfrac{2}{3}\right)\) : 2\(\dfrac{1}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{22}{15}\) : \(\dfrac{11}{5}\)
= \(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
( 2\(\dfrac{1}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(x\)) = \(\dfrac{3}{4}\)
\(\dfrac{11}{5}\) + \(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{3}{4}\) - \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{29}{20}\)
\(x\) = -\(\dfrac{29}{12}\)
6 tháng 8 2022
a: \(\Leftrightarrow\left(\dfrac{13}{4}:x\right)\cdot\left(-\dfrac{5}{4}\right)=\dfrac{-10}{6}-\dfrac{5}{6}=\dfrac{-15}{6}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{13}{4}:x=\dfrac{5}{2}\cdot\dfrac{5}{4}=\dfrac{25}{8}\)
hay \(x=\dfrac{13}{4}:\dfrac{25}{8}=\dfrac{13}{4}\cdot\dfrac{8}{25}=\dfrac{26}{25}\)
b: \(\Leftrightarrow\dfrac{3}{4}:x=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{2}{36}=\dfrac{1}{18}\)
=>\(x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{54}{4}=\dfrac{27}{2}\)
c: \(\Leftrightarrow\left(-\dfrac{6}{5}+x\right):\left(-3.6\right)=-\dfrac{7}{4}+\dfrac{1}{4}\cdot8=\dfrac{1}{4}\)
=>x-6/5=-9/10
=>x=3/10
7 tháng 11 2021
\(\Leftrightarrow x\cdot\dfrac{12}{7}=\dfrac{15}{6}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{12}{7}=\dfrac{5}{2}\cdot\dfrac{7}{12}=\dfrac{35}{24}\)
21 tháng 7 2017
Câu 1: \(\frac{5}{7}\):x+\(\frac{2}{3}\)=1\(\frac{1}{3}\)
\(\frac{5}{7}\):x+\(\frac{2}{3}\)=\(\frac{4}{3}\)
\(\frac{5}{7}\):x=\(\frac{4}{3}\) - \(\frac{2}{3}\)
\(\frac{5}{7}\):x=\(\frac{2}{3}\)
x=\(\frac{5}{7}\):\(\frac{2}{3}\)
x=\(\frac{15}{14}\)
Câu 2:x*\(\frac{3}{7}\)-\(\frac{5}{14}\)=2\(\frac{1}{3}\)
x*\(\frac{3}{7}\)-\(\frac{5}{14}\)=\(\frac{7}{3}\)
x*\(\frac{3}{7}\)=\(\frac{7}{3}\)+\(\frac{5}{14}\)
x*\(\frac{3}{7}\)=\(\frac{113}{42}\)
x=\(\frac{113}{42}\):\(\frac{3}{7}\)
x=\(\frac{791}{126}\)
Câu 3: 3\(\frac{5}{7}\)-x*\(\frac{1}{3}\)=1\(\frac{13}{11}\)
\(\frac{26}{7}\)-x*\(\frac{1}{3}\)=\(\frac{24}{11}\)
x*\(\frac{1}{3}\)=\(\frac{26}{7}\)-\(\frac{24}{11}\)
x*\(\frac{1}{3}\)=\(\frac{118}{77}\)
x=\(\frac{118}{77}\):\(\frac{1}{3}\)
x=\(\frac{356}{77}\)
\(2,4:\left(-\frac{1}{2}-x\right)=1\frac{3}{5}\)
<=> \(-\frac{1}{2}-x=\frac{3}{2}\)
<=> \(x=-2\)
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