Cho M = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\) ; N = \(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\).
Tính M \(\cdot\) N.
K
Khách
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Những câu hỏi liên quan
27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
24 tháng 3 2018
Ta có:\(A=\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{98}{99}\)
\(A< \dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{99}{100}\)
\(\Rightarrow A^2< \dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot...\cdot\dfrac{98}{99}\cdot\dfrac{99}{100}\)
\(A^2< \dfrac{2}{100}=\dfrac{1}{50}\)
Mà \(\dfrac{1}{50}< \dfrac{1}{49}\)
\(\Rightarrow A^2< \dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{7}\left(đpcm\right)\)
AH
Akai Haruma
Giáo viên
15 tháng 11 2017
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)
Xét mẫu số:
\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)
\(=2^{32}(1.2.3....31.32)\)
Suy ra:
\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)
Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)
\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)
Vậy \(x=\frac{-37}{2}\)
QN
9 tháng 5 2021
\(a.\)
\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)
\(\Leftrightarrow?=-3\)
\(b.\)
\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)
\(\Leftrightarrow?=-2\)
\(c.\)
\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow?=10\)
9 tháng 5 2021
Mk gọi ? = x nha
a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\)
\(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\)
\(\dfrac{x}{4}=\dfrac{-3}{4}\)
⇒x=-3
b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)
\(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\)
\(\dfrac{x}{3}=\dfrac{-2}{3}\)
⇒x=-2
c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\)
\(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\)
\(\dfrac{3}{x}=\dfrac{3}{10}\)
⇒x=10
14 tháng 5 2017
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
DT
Đỗ Thanh Hải
CTVVIP
14 tháng 5 2017
Ta có
A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)
A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)
A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)
A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)
A = \(\dfrac{-1}{1}\)
A = -1
Vậy A = -1
MD
11 tháng 8 2018
Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)
Ta có : M . N = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{100}{101}\)
= \(\dfrac{1}{101}\)
Vậy M . N = \(\dfrac{1}{101}\)