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8xn-1 .(\(\dfrac{1}{2}\)xn-1 - \(\dfrac{3}{4}\)x)
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Những câu hỏi liên quan
LM
Lê Minh Vũ
CTVHS
5 tháng 6 2023
\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)
= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)
= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)
= \(\dfrac{40}{14}=\dfrac{20}{7}\)
\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)
=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)
=\(\dfrac{9}{2}+\dfrac{1}{11}\)
=\(\dfrac{101}{22}\)
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)
\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)
\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)
\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)
\(x=\dfrac{102}{21}=\dfrac{34}{7}\)
25 tháng 9 2024
Mới thế đã hai năm trôi qua,câu trả lời từ mọi người vẫn KO XUẤT HIỆN.
Ko biết sau này câu trả lời có xuất hiện hay ko...
O
2 tháng 4 2022
\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)
\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)
\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)
\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)
\(C=\dfrac{5}{8}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
6 tháng 8 2023
\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{3+2}{6}-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{1}{5}=\dfrac{5\times5-6}{30}=\dfrac{19}{30}\\ \dfrac{1}{5}:4+\dfrac{3}{4}=\dfrac{1}{5}\times4+\dfrac{3}{4}=\dfrac{4}{5}+\dfrac{3}{4}=\dfrac{4\times4+3\times5}{20}=\dfrac{31}{20}\)
\(\dfrac{4}{3}\times\dfrac{9}{5}-\dfrac{3}{10}=\dfrac{36}{15}-\dfrac{3}{10}=\dfrac{12}{5}-\dfrac{3}{10}=\dfrac{12\times2-3}{10}=\dfrac{21}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 8 2023
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)
= \(\dfrac{1\times15}{2\times15}\) + \(\dfrac{1\times10}{3\times10}\) - \(\dfrac{1\times6}{5\times6}\)
= \(\dfrac{15}{30}\) + \(\dfrac{10}{30}\) - \(\dfrac{6}{30}\)
= \(\dfrac{19}{30}\)
\(\dfrac{1}{5}\) : 4 + \(\dfrac{3}{4}\)
= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)
= \(\dfrac{1}{20}\) + \(\dfrac{3\times5}{4\times5}\)
= \(\dfrac{1}{20}+\dfrac{15}{20}\)
= \(\dfrac{16}{20}\)
= \(\dfrac{4}{5}\)
\(\dfrac{4}{3}\) \(\times\) \(\dfrac{9}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{12}{5}\) - \(\dfrac{3}{10}\)
= \(\dfrac{24}{10}\) - \(\dfrac{3}{10}\)
= \(\dfrac{21}{10}\)
10 tháng 9 2021
\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)
\(=\dfrac{-5}{4}\)
12 tháng 8 2023
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
4 tháng 2 2021
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
\(8x^{n-1}.\left(\dfrac{1}{2}x^{n-1}-\dfrac{3}{4}x\right)\)
\(=4x^{2n-2}-6x^n\)
\(=4x^n.\left(x^{n-2}-2x^n\right)\)
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