cho M=$\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}$ N=$\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}$ chứng minh rằng: M

0

NV

Nguyễn Vũ Hoàng

5 tháng 11 2017

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.

1

LT

Lương Thị Vân Anh

CTVHS

5 tháng 9 2022

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 = $\dfrac{1}{101}$

DT

Duat Tran

23 tháng 3 2018

cho A=$\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{98}{99}$

Chứng minh rằng A$< \dfrac{1}{7}$

1

PN

Phạm Nguyễn Tất Đạt

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 $\dfrac{1}{50}< \dfrac{1}{49}$

$\Rightarrow A^2< \dfrac{1}{49}$

$\Rightarrow A< \dfrac{1}{7}\left(đpcm\right)$

GL

Giao Lê Nguyễn

30 tháng 3 2018

...

Đọc tiếp

1

NL

Nguyễn Lê Phước Thịnh

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: $B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}$

DD

duongmko60 đỗ

25 tháng 7 2018

Cho C=$\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\cdot\cdot\dfrac{199}{200}$ Chứng minh C²<$\dfrac{1}{201}$

2

MD

Ma Đức Minh

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)

ND

Natsu Dragneel

11 tháng 8 2018

good luck

TK

Thuy Khuat

29 tháng 10 2017

Tìm x, biết

$\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}$=$4^x$

1

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 $x=\frac{-37}{2}$

TN

Trần Ngọc Nhi

18 tháng 11 2017

Số 2 nó ở đâu chui ra v Violympic toán 7

HD

Học đi

6 tháng 12 2017

1.Tìm GTLN của biểu thức :

$\dfrac{a^{2012}+2013}{a^{2012}+2011}$

2.Cho $B=\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\cdot\cdot\dfrac{99}{100}$

CM : $\dfrac{1}{15}< B< \dfrac{1}{10}$

Làm giúp mk nha chìu mai học r !!!

0

HK

Hoàng Khánh Linh

16 tháng 3 2022

e)$\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}$

f)$\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}$
g)
$\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}$
Mn giúp em với ạ

1

HT

Hồ Tuyết Anh

VIP

16 tháng 3 2022

$e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}$

$=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}$

$=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}$

$f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}$

$=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}$

$=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}$

$g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}$

$=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)$

$=\dfrac{5}{9}\cdot1=\dfrac{5}{9}$

Đúng(3)

LY

Lê Yến

21 tháng 6 2018

$\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}..........\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2$^x

Tính giá trị các biểu thức sau theo cách hợp lí nhất. a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$; b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}}...

0

TD

Thầy Đức Anh

Giáo viên VIP

27 tháng 4 2022

Đọc tiếp

12

PT

Phạm Trường Giang

VIP

22 tháng 6 2022

a) $A = [\frac{2}{7} (\frac{1}{4} - \frac{1}{3})] : [\frac{2}{7} (\frac{1}{3} - \frac{2}{5})] = (\frac{1}{4} - \frac{1}{3}) : (\frac{1}{3} - \frac{2}{5}) = 1 \frac{1}{4}$ .
b) $B = \frac{\frac{3}{4} (\frac{1}{5} - \frac{2}{7} - \frac{1}{3} + \frac{2}{7})}{\frac{1}{5} (\frac{2}{7} + \frac{1}{3}) - \frac{1}{3} (\frac{2}{7} + \frac{1}{3})} = \frac{\frac{3}{4} (\frac{1}{5} - \frac{1}{3})}{(\frac{1}{5} - \frac{1}{3}) (\frac{2}{7} + \frac{1}{3})} = 1 \frac{11}{52}$ .

T

TINH

13 tháng 7 2022

a) $\mathrm{A}=\left[\dfrac{2}{7}\left(\dfrac{1}{4}-\dfrac{1}{3}\right)\right]:\left[\dfrac{2}{7}\left(\dfrac{1}{3}-\dfrac{2}{5}\right)\right]=\left(\dfrac{1}{4}-\dfrac{1}{3}\right):\left(\dfrac{1}{3}-\dfrac{2}{5}\right)=1 \dfrac{1}{4}$ .
b) $\mathrm{B}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{2}{7}-\dfrac{1}{3}+\dfrac{2}{7}\right)}{\dfrac{1}{5}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)-\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=\dfrac{\dfrac{3}{4}\left(\dfrac{1}{5}-\dfrac{1}{3}\right)}{\left(\dfrac{1}{5}-\dfrac{1}{3}\right)\left(\dfrac{2}{7}+\dfrac{1}{3}\right)}=1 \dfrac{11}{52}$