\(\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}-...-\dfrac{92}{100}}{\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}}\)
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21 tháng 9 2023
tỉ số của a / b là (92 - 1/9 - 2/ 10 - 3/11 - ... - 92/100) trên 1/45 + 1/50 + ... + 1/500 :)) hay ngắn tắc hơn là A/B cho nhanh :)))))))))))))))
22 tháng 9 2023
\(A=\left(1+1+...+1\right)-\left(\dfrac{1}{9}+\dfrac{2}{10}+...+\dfrac{92}{100}\right)\)𝓒𝓸́ 92 𝓼𝓸̂́ 1
\(A=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
\(A=\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
\(A=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
\(B=\dfrac{1}{45}+\dfrac{1}{50}+...+\dfrac{1}{500}\)
\(B=\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}\\ \Rightarrow\dfrac{A}{B}=\dfrac{8}{\dfrac{1}{5}}=40\)
𝓥𝓪̣̂𝔂 𝓽𝓲̉ 𝓼𝓸̂́ 𝓬𝓾̉𝓪 𝓐 𝓿𝓪̀ 𝓑 𝓵𝓪̀ 40
17 tháng 7 2017
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
31 tháng 3 2018
Cái này mk từng làm nhưng có chút sai sót vậy nên bn sữa cho mk chút nhé ! Thay vì N = ... thì bn thay bằng A = ... nha
Ta có :
N = 40 ( A = 40 )
11 tháng 6 2018
+)Đặt A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
A= \(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\left(1+1+1+...+1\right)\) (99 chữ số 1)
A= \(\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
A= \(\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+1\)
A= \(100.\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
⇒ M= \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)
M= \(\dfrac{100.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
M= 100 (1)
+) Đặt B= \(92-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\)
B= \(\left(1+1+1+...+1\right)-\dfrac{1}{9}-\dfrac{2}{10}-...-\dfrac{92}{100}\) ( 92 chữ số 1)
B= \(\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
B= \(\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
B= \(8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
⇒ N= \(\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
N= 8 (2)
Từ (1) và (2)⇒ \(\dfrac{M}{N}\) = \(\dfrac{100}{8}\)= \(\dfrac{25}{2}\)
Vậy \(\dfrac{M}{N}=\dfrac{25}{2}\)
Q
1 tháng 4 2017
Đặt vế đầu là A, vế sau là B.
Vế A:
- Tử:
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Vậy:
\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)
Vế B:
- Tử:
\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)
Vậy:
\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)
Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)
DN
28 tháng 12 2017
a) Ta có:
\(\overline{abcdeg}=10000.\overline{ab}+100.\overline{cd}+eg=9999.\overline{ab}+99.\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)\(9999.\overline{ab}⋮11\)
\(99.\overline{cd}⋮11\)
\(\overline{ab}+\overline{cd}+\overline{eg}⋮11\)
\(\Rightarrow9999.\overline{ab}+99.\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)hay \(\overline{abcdeg}⋮11\)(đpcm)
b) Ta có:
\(E=92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...\left(1-\dfrac{92}{100}\right)=\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{100}=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)\(F=\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)\)
\(\dfrac{E}{F}=\dfrac{8\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{100}\right)}=\dfrac{8}{\dfrac{1}{5}}=40\)