Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\) \(\dfrac{1}{3^2}\) \(\dfrac{1}{4^2}\) \(\dfrac{1}{5^2}\) \(\dfrac{1}{6^2}\) \(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)

1

NL

Nguyễn Lê Phước Thịnh

20 tháng 3 2023

1:

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)

=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

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NH

Nguyễn Hoàng Gia Bảo

2 tháng 5 2021

Chứng tỏ rằng B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}< 1\)

1

DT

Đỗ Thanh Hải

CTVVIP

2 tháng 5 2021

Ta có

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

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

\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)

=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)

B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))

Vậy.....

\(\text{Bài 4. Chứng tỏ rằng:}\)\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17} 2\)\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}...

E

ᴵᴬᴹɴɢᴜʏễɴ●ɴɢọᴄ●ʜùɴɢッ

11 tháng 4 2021

2

NA

Ngoc Anh Thai

Giáo viên

11 tháng 4 2021

a)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)

b)

\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)

NA

Ngoc Anh Thai

Giáo viên

11 tháng 4 2021

c)

\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)

d) tương tự câu 1

LT

Lê Thị Bích

12 tháng 4 2017

Chứng tỏ rằng: B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2} +\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)<1

Bài 2: (Đề1) Tính \(4\dfrac{2}{3}+3\dfrac{2}{7}\)=.......................... \(8\dfrac{5}{9}:5\dfrac{1}{2}\) =.......................... \(6\dfrac{5}{7}:2\dfrac{1}{6}\) =.......................... ...

2

KT

Không Thể Nói

12 tháng 4 2017

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)

mink nhanh nhất đó bạn,

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NM

Nguyễn Mai Phương

4 tháng 5 2018

ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1\times2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\times3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\times4}\)

. . . . . . .

\(\dfrac{1}{8^2}< \dfrac{1}{7\times8}\)

_________________________________

\(\Rightarrow\)\(B< \)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{7}-\dfrac{1}{8}\)

\(\Rightarrow B< 1-\dfrac{1}{8}\)

\(\Rightarrow B< 1\)

\(\Rightarrowđpcm\)

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MG

Mai gia bảo

31 tháng 7 2023

Bài 1. Tính A= \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) B= \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)Bài 2. Tính a) \(5\dfrac{1}{2}.3\dfrac{1}{4}\) b) \(6\dfrac{1}{3}:4\dfrac{2}{9}\) ...

#Toán lớp 5

1

ND

Nguyễn Đức Trí

VIP

31 tháng 7 2023

\(4\dfrac{2}{3}+3\dfrac{2}{7}=\dfrac{14}{3}+\dfrac{23}{7}=\dfrac{14x7+23x3}{21}=\dfrac{167}{21}\)

\(8\dfrac{5}{9}:5\dfrac{1}{2}=\dfrac{77}{9}:\dfrac{11}{2}=\dfrac{77}{9}x\dfrac{2}{11}=\dfrac{14}{9}\)

\(6\dfrac{5}{7}:2\dfrac{1}{6}=\dfrac{47}{7}:\dfrac{13}{6}=\dfrac{47}{7}x\dfrac{6}{13}=\dfrac{282}{91}\)

\(1\dfrac{3}{4}x2\dfrac{5}{6}=\dfrac{7}{4}x\dfrac{17}{6}=\dfrac{119}{24}\)

\(5\dfrac{3}{4}-2=\dfrac{23}{4}-2=\dfrac{23}{4}-\dfrac{8}{4}=\dfrac{15}{4}\)

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MN

Miru nèe

16 tháng 6 2021

3

Y

Yeutoanhoc

16 tháng 6 2021

`A=(8 2/7-4 2/7)-3 4/9`

`=8+2/7-4-2/7-3-4/9`

`=4-3-4/9`

`=1-4/9=5/9`

`B=(10 2/9-6 2/9)+2 3/5`

`=10+2/9-6-2/9+2+3/5`

`=4+2+3/5`

`=6+3/5=33/5`

Bài 2:

`a)5 1/2*3 1/4`

`=11/2*13/4`

`=143/8`

`b)6 1/3:4 2/9`

`=19/3:38/9`

`=19/3*9/38=3/2`

`c)4 3/7*2`

`=31/7*2`

`=62/7`

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T

꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...

16 tháng 6 2021

Bài 1:

\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)

\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)

\(A=4-\dfrac{31}{9}\)

\(A=\dfrac{5}{9}\)

\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)

\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)

\(B=4+\dfrac{13}{5}\)

\(B=\dfrac{33}{5}\)

TT

Trần Thảo Nguyên

13 tháng 7 2023

Bài 5:

a) 4\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) : 5\(\dfrac{1}{2}\)

b) \(\dfrac{7}{6}\) : (\(\dfrac{1}{2}\) x \(\dfrac{3}{4}\)) - \(\dfrac{5}{8}\)

#Toán lớp 5

1

ND

Nguyễn Đức Trí

VIP

13 tháng 7 2023

a) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}x\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)

b) \(\dfrac{7}{6}:\left(\dfrac{1}{2}x\dfrac{3}{4}\right)-\dfrac{5}{8}=\dfrac{7}{6}:\dfrac{3}{8}-\dfrac{5}{8}=\dfrac{7}{6}x\dfrac{8}{3}-\dfrac{5}{8}=\dfrac{28}{9}-\dfrac{5}{8}=\dfrac{224}{72}-\dfrac{45}{72}=\dfrac{179}{72}\)

MG

Mai gia bảo

3 tháng 8 2023

Bài 1: (đề 2) Tính

a) \(2\dfrac{3}{8}:3\dfrac{3}{4}x2\dfrac{2}{3}\)=............

b) \(1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\)=............

c) \(4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\)=............

#Toán lớp 5

2

ND

Nguyễn Đức Trí

VIP

3 tháng 8 2023

a) \(...=\dfrac{19}{8}:\dfrac{15}{4}x\dfrac{8}{3}=\dfrac{19}{8}x\dfrac{4}{15}x\dfrac{8}{3}=\dfrac{76}{45}\)

b) \(...=\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{17}{6}=\dfrac{3}{2}x\dfrac{3}{7}x\dfrac{6}{17}=\dfrac{27}{119}\)

c) \(...=\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}=\dfrac{14}{3}-\dfrac{7}{4}x\dfrac{5}{12}=\dfrac{14}{3}-\dfrac{35}{48}=\dfrac{14x16}{48}-\dfrac{35}{48}=\dfrac{224}{48}-\dfrac{35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

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GN

GV Nguyễn Trần Thành Đạt

Giáo viên

3 tháng 8 2023

\(a,2\dfrac{3}{8}:3\dfrac{3}{4}\times2\dfrac{2}{3}\\ =\dfrac{2\times8+3}{8}:\dfrac{3\times4+3}{4}\times\dfrac{2\times3+2}{3}\\ =\dfrac{19}{8}:\dfrac{15}{4}\times\dfrac{8}{3}\\ =\dfrac{19\times4\times8}{8\times15\times3}=\dfrac{76}{45}\)

\(b,1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\\ =\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{2\times6+5}{6}\\ =\dfrac{3}{2}\times\dfrac{3}{7}\times\dfrac{6}{17}\\ =\dfrac{54}{238}=\dfrac{27}{119}\)

\(c,4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\\ =\dfrac{4\times3+2}{3}-\dfrac{1\times4+3}{4}:\dfrac{2\times5+2}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}.\dfrac{5}{12}\\ =\dfrac{14}{3}-\dfrac{35}{48}\\ =\dfrac{14\times16-35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

22 tháng 2 2022

a, Tính: M = \(1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)

b, Chứng tỏ: S = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(n\in N,n\ge2\right)\)