Bài 1: Chứng minh rằng tổng S= $\dfrac{1}{2^2}$- $\dfrac{1}{2^4}$ $\dfrac{1}{2^6}$- ... $\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$ ... $\dfrac{1}{2^{2018}}$-$\dfrac{1}{2^{2020}}$

ND

Ngọc Diễm

27 tháng 6 - olm

Bài 1: Chứng minh rằng tổng S= $\dfrac{1}{2^2}$- $\dfrac{1}{2^4}$+$\dfrac{1}{2^6}$- ...+ $\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$+... + $\dfrac{1}{2^{2018}}$-$\dfrac{1}{2^{2020}}$ <0,2 Bài 2: Cho T= 22020- 22019-22018-...-2-1. Tính 2021T Bài 3: Tìm số tự nhiên x, biết $\dfrac{7^{x+2}+7^{x+1}+7^x}{57}$=$\dfrac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}$ Bài 4: Tìm x, biết a) $\dfrac{1}{4}$. $\dfrac{2}{6}$. $\dfrac{3}{8}$... $\dfrac{30}{62}$. $\dfrac{31}{64}$=...

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#Toán lớp 7

2

H9

HT.Phong (9A5)

CTVHS

27 tháng 6

Bài 4:

a)

$\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=4^x\\ \Rightarrow\dfrac{1\cdot2\cdot3\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot...\cdot62\cdot64}=4^x\\ \Rightarrow\dfrac{2\cdot3\cdot4\cdot...\cdot30\cdot31}{\left(2\cdot2\right)\cdot\left(3\cdot2\right)\cdot\left(4\cdot2\right)\cdot...\cdot\left(2\cdot31\right)\cdot\left(2\cdot32\right)}=\left(2^2\right)^x\\ \Rightarrow\dfrac{2\cdot3\cdot4\cdot...\cdot30\cdot31}{2^{31}\cdot\left(2\cdot3\cdot4\cdot...\cdot31\cdot32\right)}=2^{2x}\\ \Rightarrow\dfrac{1}{2^{31}\cdot32}=2^{2x}\\ \Rightarrow\dfrac{1}{2^{31}\cdot2^5}=2^{2x}\\ \Rightarrow\dfrac{1}{2^{36}}=2^{2x}\\ \Rightarrow2^{-36}=2^{2x}\\ \Rightarrow2x=-36\\ \Rightarrow x=-18$

b)

$\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\\ \Rightarrow\dfrac{4\cdot4^5}{3\cdot3^5}\cdot\dfrac{6\cdot6^5}{2\cdot2^5}=8^x\\ \Rightarrow\dfrac{4^6}{3^6}\cdot\dfrac{6^6}{2^6}=8^x\\ \Rightarrow\dfrac{4^6}{3^6}\cdot3^6=8^x\\ \Rightarrow4^6=8^x\\ \Rightarrow\left(2^2\right)^6=\left(2^3\right)^x\\ \Rightarrow2^{12}=2^{3x}\\ \Rightarrow3x=12\\ \Rightarrow x=4$

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NL

Ngô Lê Kim Nhi

27 tháng 6

Làm seo để viết phân số thế?

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HD

Hải Dương

9 tháng 7 2017

Chứng minh$B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}>\dfrac{1}{2004}$

Chứng minh $S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004 }}< 0.2$

#Toán lớp 7

1

NT

Nguyễn Thanh Hằng

9 tháng 7 2017

a)

$B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...........-\dfrac{1}{2004^2}$

$\Leftrightarrow B=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..............+\dfrac{1}{2004^2}\right)$

Đặt :

$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.............+\dfrac{1}{2004^2}$

Ta thấy :

$\dfrac{1}{2^2}< \dfrac{1}{1.2}$

$\dfrac{1}{3^2}< \dfrac{1}{2.3}$

..........................

$\dfrac{1}{2004^2}< \dfrac{1}{2003.2004}$

$\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+..............+\dfrac{1}{2003.2004}$

$\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{2003}-\dfrac{1}{2004}$

$\Leftrightarrow A< 1-\dfrac{1}{2004}$

$\Leftrightarrow A< \dfrac{2003}{2004}$

$\Leftrightarrow1-A< 1-\dfrac{2003}{2004}$

$\Leftrightarrow B< \dfrac{1}{2004}\left(đpcm\right)$

b) $S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-........+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}$

$\Leftrightarrow2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.....+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+....+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$

$\Leftrightarrow4S=1-\dfrac{1}{2^2}+.......+\dfrac{1}{2^{4n}}-\dfrac{1}{2^{4n+2}}+.......+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}$

$\Leftrightarrow4S+S=\left(1-\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+.......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$$\Leftrightarrow5S=1-\dfrac{1}{2^{2004}}< 1$

$\Leftrightarrow S< \dfrac{1}{5}=0,2$

$\Leftrightarrow S< 0,2\left(đpcm\right)$

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TT

thao truong

19 tháng 2 2020

cho mik hỏi mik ko hiểu tại sao từ 1/2^4n-2 khi nhân với 2^2 lại ra đc 1/2^4n vậy? Xin hãy giải đáp giùm mik

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HT

Hà Thúy Phương

3 tháng 3 2019

Chứng minh rằng tổng :

S=$\dfrac{1}{2^2}$- $\dfrac{1}{2^4}$+$\dfrac{1}{2^6}$-...+$\dfrac{1}{2^{4n-2}}$-$\dfrac{1}{2^{4n}}$+...+$\dfrac{1}{2^{2002}}$-$\dfrac{1}{2^{2004}}$<0.2

#Toán lớp 7

1

S

svtkvtm

3 tháng 3 2019

$S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\Rightarrow4S=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+......-\dfrac{1}{2^{2002}}\Rightarrow4S+S=5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< 0,2\left(\text{đpcm}\right)$

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DN

Dương Ngọc Ánh

6 tháng 12 2023 - olm

chứng minh rằng: $\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}$

#Toán lớp 6

1

AH

Akai Haruma

Giáo viên

6 tháng 12 2023

Lời giải:

Đặt $A=\frac{1}{7^2}-\frac{1}{7^4}+....+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}$

$7^2A=1-\frac{1}{7^2}+....+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n-2}}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}$

$\Rightarrow A+7^2A=1-\frac{1}{7^{100}}\Rightarrow 50A=1-\frac{1}{7^{100}}<1$

$\Rightarrow A< \frac{1}{50}$

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LP

Lương Phạm

6 tháng 4 2021

$CMR,S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{20002}}-\dfrac{1}{2^{2004}}< 0,2$

#Toán lớp 7

0

TD

Thái Đào

13 tháng 3 2017

CM

$S=\dfrac{1}{2^2}-\dfrac{1}{2^4}-\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}< 0,2$

#Toán lớp 7

1

LF

Lightning Farron

13 tháng 3 2017

$S=\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}$

$2^2S=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$

$4S=1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{4n}}+\dfrac{1}{2^{4n+2}}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}$

$4S+S=\left(1-\dfrac{1}{2^2}+...+\dfrac{1}{2^{2000}}-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$

$5S=1-\dfrac{1}{2^{2004}}< 1\Rightarrow S< \dfrac{1}{5}=0,2$

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TT

Trịnh Thị Thảo Nhi

27 tháng 11 2017

Chứng minh rằng:

$\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}< \dfrac{1}{50}$

#Toán lớp 7

1

LT

Linh Trần

27 tháng 11 2017

Đặt $A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}$

Ta có:

$\dfrac{A}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}$

$\Rightarrow A+\dfrac{A}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\right)$

$\Rightarrow\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}$

$\Rightarrow A< \dfrac{1}{50}$

=> ĐPCM.

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NH

Nguyễn Huyền Trâm

13 tháng 10 2019

Chứng minh rằng :

$\dfrac{1}{7^2} - \dfrac{1}{7^4} + ... + \dfrac{1}{7^{4n-2}} - \dfrac{1}{7^{4n}} + ... + \dfrac{1}{7^{98}}-\dfrac{1}{7^{100}} < \dfrac{1}{50}$

#Toán lớp 7

1

S

svtkvtm

13 tháng 10 2019

$\text{Đặt:}S=\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\Rightarrow49S=1-\frac{1}{7^2}+.....-\frac{1}{7^{98}}\Rightarrow49S+S=50S=\left(1-\frac{1}{7^2}+\frac{1}{7^4}-....-\frac{1}{7^{98}}\right)+\left(\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\right)=1-\frac{1}{7^{100}}< 1\Rightarrow S< \frac{1}{50}\left(\text{đpcm}\right)$

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NH

Nguyễn Huyền Trâm

13 tháng 10 2019

svtkvtm mơn bn nhìu nhìu

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NH

nguyễn hà

20 tháng 3 2018

$\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+...+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}$<0,2

#Toán lớp 7

1

NT

Nguyễn Thanh Hằng

20 tháng 3 2018

Đặt :

$A=\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+.........+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}$

$\Leftrightarrow2^2A=2^2\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+\dfrac{1}{2^6}-.......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$

$\Leftrightarrow4A=1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-.......-\dfrac{1}{2^{2002}}$

$\Leftrightarrow4A+A=\left(1-\dfrac{1}{2^2}+\dfrac{1}{2^4}-\dfrac{1}{2^6}+.......-\dfrac{1}{2^{4n-2}}+\dfrac{1}{2^{4n}}-......-\dfrac{1}{2^{2002}}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^4}+......+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}+......+\dfrac{1}{2^{2002}}-\dfrac{1}{2^{2004}}\right)$

$\Leftrightarrow5A=1-\dfrac{1}{2^{2004}}$

$\Leftrightarrow A=\left(1-\dfrac{1}{2^{2004}}\right):5$

$\Leftrightarrow A=\dfrac{1}{5}-\dfrac{1}{5}.\dfrac{1}{2^{2004}}< \dfrac{1}{5}=0,2\left(đpcm\right)$

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NH

nguyễn hà

20 tháng 3 2018

mk cám ơn nha

lm bn nhé