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\(\dfrac{2}{7}\) +\(\dfrac{3}{2}\) + \(\dfrac{1}{3}\)
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10 tháng 3 2022
\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)
HN
10 tháng 3 2022
1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.
2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.
3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).
8 tháng 1 2021
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
16 tháng 6 2021
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
16 tháng 6 2021
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
8 tháng 7 2023
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
Y
6 tháng 6 2023
\(\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{23}}\times\dfrac{\dfrac{1}{3}-0,25-0,2}{1\dfrac{1}{6}-0,875-0,7}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{23}-\dfrac{1}{23}}\times\dfrac{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}-\dfrac{7}{10}}\)
\(=\dfrac{\dfrac{1}{3} +\dfrac{1}{7}-\dfrac{1}{23}}{\dfrac{1}{3}\times2+\dfrac{1}{7}\times2-\dfrac{1}{23}\times2}\times\dfrac{\dfrac{2}{6}-\dfrac{2}{8}-\dfrac{2}{10}}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}}{2\times\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{23}\right)}\times\dfrac{2\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}{7\times\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\)
\(=\dfrac{1}{2}\times\dfrac{2}{7}\)
\(=\dfrac{1}{7}\)
27 tháng 6 2023
d: =-2/7-3/11+2/7=-3/11
e: =2+3/7+1+4/7-17/7
=4-17/7=11/7
f: =-2/3*4/5+1/5*11/9
=-8/15+11/45
=-24/45+11/45=-13/45
h: =(-3,1+3,7):0,2=0,6:0,2=3
MH
9 tháng 5 2022
Bài 1:
+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)
\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)
+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)
\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)
\(y=\dfrac{9}{50}\)
12 tháng 3 2023
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
\(=\dfrac{12}{42}+\dfrac{63}{42}+\dfrac{14}{42}=\dfrac{89}{42}\)
\(\dfrac{2}{7}\) + \(\dfrac{3}{2}\) + \(\dfrac{1}{3}\) = \(\dfrac{12}{42}\) + \(\dfrac{63}{42}\) + \(\dfrac{14}{42}\) = \(\dfrac{89}{42}\)