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A=\(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}-...-\dfrac{1}{2^{58}}\)
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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:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
LM
Lê Minh Vũ
VIP
5 tháng 6 2023
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
Sửa đề: \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\)
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+.....+\dfrac{1}{2^{58}}\)
\(\Rightarrow2^3A=2^3.\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+\dfrac{1}{2^{58}}\right)\)
\(\Rightarrow2^3A=1+\dfrac{1}{2}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{55}}\)
\(\Rightarrow2^3A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^7}+....+\dfrac{1}{2^{58}}\right)\)\(\Rightarrow7A=1-\dfrac{1}{2^{58}}\Rightarrow A=\dfrac{1-\dfrac{1}{2^{58}}}{7}\)
Vậy...........
~ Học tốt nha ~
\(A=\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\\ 2^3\cdot A=\dfrac{2^3}{2}+\dfrac{2^3}{2^4}+\dfrac{2^3}{2^7}+...+\dfrac{2^3}{2^{58}}\\ 8A=4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\\ 8A-A=\left(4+\dfrac{1}{2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{55}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^4}+\dfrac{1}{2^7}+...+\dfrac{1}{2^{58}}\right)\\ 7A=4-\dfrac{1}{2^{58}}\\ A=\dfrac{4-\dfrac{1}{2^{58}}}{7}\)