a) cho A=1/1x3 1/3.5 1/5/7 1/7.9 1/9.11So sánh a với 1/2b)tính B=1/1.4 1/4.7 1/7.10 .... 1/2002.2005 1/2005.2008c)tính B=1/1.2 1/2.3 ... 1/18.19 1/19.20d)cho S=3/1.4 3/4.7 3/7.10 .... 3/40.43 1/43.46C...

NT

nguyen thi yen chi

21 tháng 9 2018

a) cho A=1/1x3+1/3.5+1/5/7+1/7.9+1/9.11So sánh a với 1/2b)tính B=1/1.4+1/4.7+1/7.10+....+1/2002.2005+1/2005.2008c)tính B=1/1.2+1/2.3+...+1/18.19+1/19.20d)cho S=3/1.4+3/4.7+3/7.10+....+3/40.43+1/43.46Chứng minh S<1e)\(tínhP=\left(1\frac{1}{2}\right).\left(1\frac{1}{3}\right).\left(1\frac{1}{4}\right)......\left(1\frac{1}{20}\right)\)bài 2:Chứng tỏ rằnga)ab+ba:11b)ba-ab\(⋮\)9(b>a)c)abc-cba chia hết 99 AI NHANH MK TICK CHO...

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#Toán lớp 6

0

IL

I LOVE KOOKIE

9 tháng 4 2017

tính

a) P = 1 / 1.2 + 2 / 2.4 + 3 / 4.7 + ...+ 10 / 46.56

b) A= 3 / 1.2 + 3 / 2.3 + 3 / 3.4 + ....+ 3 / 99.100 chú ý : / là phần nha

c) B = 3 / 1.4 + 3 / 4.7 + 3 / 7.10 + ... + 3 / 100.103

d) C= 5 / 1.4 + 5 / 4.7 + 5 / 7.10 + ...+ 5 / 100.103

e) D= 7 / 1.5 + 7 / 5.9 + 7 / 9.13 +...+ 7 / 101.105

#Toán lớp 6

1

D

DanAlex

9 tháng 4 2017

a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)

=\(1-\frac{1}{56}=\frac{55}{56}\)

b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)

= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)

=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)

=> \(A=\frac{99}{100}.3=\frac{297}{100}\)

c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)

=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)

=\(1-\frac{1}{103}=\frac{102}{103}\)

=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)

e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)

=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)

=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)

=\(1-\frac{1}{105}=\frac{104}{105}\)

=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)

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VZ

Vin Zoi

15 tháng 2 2016

Tính:

a/ -1/n - 1/n+a

b/ 1/1.2 + 1 /2.3 + 1/3.4 +......+ 1/2007.2008

c/ 3/1.4 + 3/4.7 + 3/7.10 +........+ 3/94.97

#Toán lớp 6

1

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

19 tháng 3

a; \(\dfrac{-1}{n}\) - \(\dfrac{1}{n+a}\)

= \(\dfrac{-n-a-n}{n.\left(n+a\right)}\)

= \(\dfrac{-2n-a}{n.\left(n+a\right)}\)

b; \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ....+ \(\dfrac{1}{2007.2008}\)

= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

= \(\dfrac{1}{1}\) - \(\dfrac{1}{2008}\)

= \(\dfrac{2007}{2008}\)

c; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

= \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

= \(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)

= \(\dfrac{96}{97}\)

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NA

Nguyễn Anh Vũ

2 tháng 10 2023

1.

a) 1/1.4+1/4.7+1/7.10+...+1/100.103

b)-1/3+-1/15+-1/35+-1/63+...+-1/9999

2.

3/1.4+3/4.7+3/7.10+...+3/94.97+3/97.100

#Toán lớp 7

1

G0

Grey.nnvd (07)

2 tháng 10 2023

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

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IL

I LOVE KOOKIE

9 tháng 4 2017

tính

a) P = 1 / 1.2 + 2 / 2.4 + 3 / 4.7 + ...+ 10 / 46.56

b) A= 3 / 1.2 + 3 / 2.3 + 3 / 3.4 + ....+ 3 / 99.100 chú ý : / là phần nha

c) B = 3 / 1.4 + 3 / 4.7 + 3 / 7.10 + ... + 3 / 100.103

d) C= 5 / 1.4 + 5 / 4.7 + 5 / 7.10 + ...+ 5 / 100.103

e) D= 7 / 1.5 + 7 / 5.9 + 7 / 9.13 +...+ 7 / 101.105

#Toán lớp 6

1

Q

Quìn

9 tháng 4 2017

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

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

\(A=3\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}\)

\(A=\dfrac{297}{100}\)

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)

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NT

Nguyễn Trúc

27 tháng 4 2021

a.Chứng tỏ rằng B = 1/2² + 1/3² + 1/4² + 1/5² + 1/6²+ 1/7² +1/8² < 1

b.Cho S = 3/1.4 + 3/4.7 + 3/7.10 +......+3/40.43 + 3/43.46 hãy chứng tỏ rằng S < 1

#Toán lớp 6

2

OM

✪ ω ✪Mùa⚜ hoa⚜ phượng⚜ là⚜ mùa⚜ th...

27 tháng 4 2021

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

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

\(...\)

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

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

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

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

Vậy B < 1 (đpcm)

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T

꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...

27 tháng 4 2021

Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

B<1/1-1/8

B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1/1-1/46

S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

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HL

hoang le

11 tháng 6 2015

S=3/1.4+3/4.7+3/7.10+...+3/40.43+43.46 CHứng minh S<1

#Toán lớp 6

1

KK

Kuroba Kaito

11 tháng 6 2015

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}

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LP

Lê Phương Linh

23 tháng 9 2023

...

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#Toán lớp 7

5

KT

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』

23 tháng 9 2023

`#3107`

`a)`

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}\)

\(=\dfrac{1999}{2000}\)

`b)`

\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

`c)`

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

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VN

Võ Ngọc Phương

VIP

23 tháng 9 2023

b) Sửa đề:

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

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HC

hùng châu mạnh hào

23 tháng 4 2016

Cho S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/32+46 .Hay chung to rang S<1

#Toán lớp 6

1

NH

Nguyên Hà Linh

23 tháng 4 2016

Hình như có sai đề~~~

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HD

Hoàng Dương Bảo Anh

19 tháng 4 2016

cho S= 3/1.4+3/4.7+3/7.10+....+3/40.43+3/43. 46

chứng tở rằng S<1

#Toán lớp 6

2

KZ

Kalluto Zoldyck

19 tháng 4 2016

S = 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 +.............+ 1/40 - 1/43 + 1/43 - 1/46

S = 1 - 1/46

S = 45/46 < 1

=> S < 1 (đpcm)

*** k mk nha các bạn ***

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NT

Nguyễn Thị Ánh Tuyết _291

19 tháng 4 2016

\(s=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{43.46}\)

\(\Rightarrow s=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{43}-\frac{1}{46}\)

\(\Rightarrow S=1-\frac{1}{46}=\frac{45}{46}\)

Tao có \(\frac{45}{46}<1\) => S < 1

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