Chứng minh rằng : \(\dfrac{1}{5}+\dfrac{1}{15} +\dfrac{1}{25}+...+\dfrac{1}{1985}

DV

Đức Vương Hiền

19 tháng 1 2019

Chứng minh rằng : $\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}< \dfrac{9}{20}$

#Toán lớp 7

0

CN

Cô Nàng Song Tử

17 tháng 4 2017

Chứng minh rằng: $\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+.....+\dfrac{1}{1985}< \dfrac{9}{20}$

#Toán lớp 7

1

ND

Natsu Dragneel

12 tháng 8 2018

Ta có :

$A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+...+\dfrac{1}{1985}$

$A=\dfrac{1}{5}+\dfrac{1}{3.5}+\dfrac{1}{5.5}+\dfrac{1}{7.5}+...+\dfrac{1}{397.5}$

$\Rightarrow5A=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}$

$5A-1=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}$

$5A-1=\dfrac{1}{3}+\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)+\left(\dfrac{1}{11}+\dfrac{1}{13}+...+\dfrac{1}{27}\right)+$

$\left(\dfrac{1}{29}+\dfrac{1}{31}+...+\dfrac{1}{81}\right)+\left(\dfrac{1}{83}+\dfrac{1}{85}+...+\dfrac{1}{243}\right)+...+\dfrac{1}{397}$

$\Rightarrow5A-1>\dfrac{1}{3}+\dfrac{1}{9}.3+\dfrac{1}{27}.9+\dfrac{1}{81}.27+\dfrac{1}{243}.81=\dfrac{1}{3}.5=\dfrac{5}{3}$

$\Rightarrow5A-1>\dfrac{5}{4}\Rightarrow5A>\dfrac{9}{4}$

$\Rightarrow A>\dfrac{9}{4}:5=\dfrac{9}{20}\Rightarrow\left(dpcm\right)$

Đúng(0)

TR

Trang Rabbit

4 tháng 12 2017

Thực hiện phép tính sau: a, |-0,75| + $\dfrac{1}{4}-2\dfrac{1}{2}$ b, $15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)$ c, $\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}$ d, $\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}$ giúp mk nha mk đang cần...

Đọc tiếp

#Toán lớp 7

1

DN

Đỗ Nguyễn Đức Trung

4 tháng 12 2017

a)$\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}$

=0.75+0.25-2.5

=1-2.5=-1.5

b)$15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)$

=3.(-1.4)+3.08

=-4.2+3.08=-1.12

c)$\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}$

=$\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}$

=$\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}$

=$\dfrac{11}{153}+\dfrac{12}{17}$

=$\dfrac{7}{9}$

d)$\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}$

=$\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}$

=-0.44+$\dfrac{127}{175}$

=$\dfrac{2}{7}$

Đúng(0)

HT

Hoàng Thúy An

19 tháng 9 2017

tinh nhanh

A=$\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}$

B=$\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}$

giup minh nhe minh dang can gap

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 5 2022

$A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}$

$=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}$

=1/64

Đúng(0)

HT

hoang thuy an

19 tháng 9 2017 - olm

tinh nhanh

A=\dfrac{1}{3}-\dfrac{3}{4}-\left(\dfrac{-3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}

B=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{3}+\dfrac{13}{15}+\dfrac{11}{15}-\dfrac{9}{11}+\dfrac{7}{9}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{3}-\dfrac{1}{3}

giup minh nhe minh dang can gap

#Toán lớp 7

0

NT

Nguyễn Thị Thoa

10 tháng 11 2023 - olm

Cho biểu thức A=$\dfrac{1}{2^2}$+$\dfrac{1}{3^2}$+$\dfrac{1}{4^2}$+$\dfrac{1}{5^2}$+$\dfrac{1}{6^2}$+$\dfrac{1}{7^2}$+$\dfrac{1}{8^2}$+$\dfrac{1}{9^2}$+$\dfrac{1}{10^2}$

Chứng minh rằng A<1

#Toán lớp 7

2

H9

HT.Phong (9A5)

CTVHS

10 tháng 11 2023

Ta có:

$\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}$

$\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}$

$\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}$

...

$\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}$

$\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}$

$\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}$

$\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}$

$\Rightarrow A< 1-\dfrac{1}{10}$

$\Rightarrow A< \dfrac{9}{10}$

$\Rightarrow A< 1$ (vì: $\dfrac{9}{10}< 1$)

Đúng(2)

VT

Vũ thị Tình

VIP

10 tháng 11 2023

$2 ^{2} 1 = 2 \cdot 2 1 < 1 \cdot 2 1$

$3 ^{2} 1 = 3 \cdot 3 1 < 2 \cdot 3 1$

$4 ^{2} 1 = 4 \cdot 4 1 < 3 \cdot 4 1$

...

$9 ^{2} 1 = 9 \cdot 9 1 < 8 \cdot 9 1$

$1 0 ^{2} 1 = 10 \cdot 10 1 < 9 \cdot 10 1$

$\Rightarrow A = 2 ^{2} 1 + 3 ^{2} 1 + 4 ^{2} 1 + ... + 1 0 ^{2} 1 < 1 \cdot 2 1 + 2 \cdot 3 1 + 3 \cdot 4 1 + ... + 9 \cdot 10 1$

$\Rightarrow A < 1 - 2 1 + 2 1 - 3 1 + ... + 9 1 - 10 1$

$\Rightarrow A < 1 - 10 1$

$\Rightarrow A < 10 9$

$\Rightarrow A < 1$ (vì: $10 9 < 1$ )

Đúng(1)

ND

Nguyễn Đỗ Anh Quân

26 tháng 2 2018

Chứng minh rằng: $\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2}$

giúp mk nha !! ^_^

#Toán lớp 7

1

BN

Bích Ngọc

26 tháng 2 2018

Chứng minh 1 bất đẳng thức cơ bản sau:$\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}$

Thật vậy: $\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{n^2+n^2+2n+1}=\dfrac{1}{2n^2+2n+1}< \dfrac{1}{2n^2+2n}=\dfrac{1}{2n\left(n+1\right)}$

Thay vào bài toán $\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+\dfrac{1}{3^2+\left(3+1\right)^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}$

$< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)$

$=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}-\dfrac{1}{2\left(n+1\right)}< \dfrac{1}{2}\left(đpcm\right)$

Đúng(0)

B

blabla

26 tháng 7 2023

a $\dfrac{-4}{7}$ - $\dfrac{5}{13}$ x $\dfrac{-39}{25}$ + $\dfrac{-1}{42}$ : $\dfrac{-5}{6}$b $\dfrac{2}{9}$ x [$\dfrac{4}{45}$: ( $\dfrac{1}{5}$ - $\dfrac{2}{15}$) + 1$\dfrac{2}{3}$]...

Đọc tiếp

#Toán lớp 7

1

MP

Minh Phương

26 tháng 7 2023

$a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}$

$=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}$ $=\dfrac{1}{35}+\dfrac{1}{35}=\dfrac{2}{35}$

$b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}$

$=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}$

$=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}$

$=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}$

Đúng(1)

LQ

Lê Quang Dũng

24 tháng 7 2017

9 Cho A= $\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}.$ Chứng minh rằng : $\dfrac{7}{12}< A< \dfrac{5}{6}$

#Toán lớp 7

2

MS

Mashiro Shiina

24 tháng 7 2017

$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$

$A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+..+\dfrac{1}{9900}$

$A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9900}\right)$

$A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)$

$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$

$A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$A< \dfrac{5}{6}\left(2\right)$

$\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm$

Đúng(3)

NT

Nguyễn Thanh Hằng

24 tháng 7 2017

Ta có :

$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}$

$\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+............+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$

$\Leftrightarrow A>\dfrac{1}{12}$$\left(1\right)$

Lại có :

$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...........+\dfrac{1}{99.100}$

$\Leftrightarrow A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-.........-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}$

$\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

$\Leftrightarrow A< \dfrac{5}{6}\left(2\right)$

Từ $\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm$

Đúng(0)

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP