a ) Chứng minh : $\frac{1}{1.2} \frac{1}{3.4} \frac{1}{5.6} ... \frac{1}{49.50}=\frac{1}{26} \frac{1}{27} \frac{1}{28} ... \frac{1}{50}$ b ) Cho \(A=\frac{1}{1.2} \frac{1}{3.4} \frac{1}{5.6} ... \f...

TQ

Trần Quốc Tuấn hi

2 tháng 12 2019

a ) Chứng minh : $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$ b ) Cho $A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}.cmr:\frac{7}{12} A \frac{5}{6}$ Các bạn giúp mình nhé : Bạn Vũ Minh Tuấn , Nguyễn Việt Lâm , Nguyễn Văn Đạt , Băng Băng 2k6 và thầy Akai Haruma , Phynit và cùng với tất cả các bạn khác vào giúp mình với ạ...

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#Toán lớp 7

0

X

Xmaf

27 tháng 3 2019 - olm

Chứng minh rằng l

$A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

2

ZC

zZz Cool Kid_new zZz

27 tháng 3 2019

$A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)$

$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.....+\frac{1}{50}^{ĐPCM}$

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NV

Nguyễn Việt Hoàng

27 tháng 3 2019

$A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)$

$A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

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LN

Lê Nguyễn Ngân Anh

25 tháng 7 2016 - olm

Chứng minh rằng:$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

3

VA

van anh ta

25 tháng 7 2016

Ta có :

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)$

$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

=> đpcm

Ủng hộ mk nha !!! ^_^

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S

Sarah

30 tháng 7 2016

$\text{Ta có :}$$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)$

$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

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TN

Thảo Nhi Nguyễn Trần

12 tháng 10 2018 - olm

Chứng minh rằng:

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

3

NV

๛Ňɠũ Vị Čáէツ

12 tháng 10 2018

Ta có :

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2.\frac{1}{2}-2.\frac{1}{4}-2.\frac{1}{6}-...-2.\frac{1}{50}$

$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}$

$=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}$

Vậy ......

~Hok tốt~

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PD

Phạm Đôn Lễ

12 tháng 10 2018

CỨ TÍNH VẾ TRÁI R ĐC

1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

(1+1/3+1/5+...+1/49)+(1+1/2+1/4+...+1/50)-(1/2+1/4+1/6+...+1/50)-(1/2+1/4+1/6+...+1/50)

(1+1/2+1/3+1/4+...+1/50)-2(1/2+1/4+1/6+..+1/50)

(1+1/2+1/4+1/6+...+1/50)-(1+1/2+1/3+..+1/25)

1/26+1/27+...+1/50

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TM

Trần Minh Anh

15 tháng 1 2017

Chứng minh rằng:

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

1

PT

Phương Trâm

15 tháng 1 2017

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}$

$=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{25}\right)$

$=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\left(đpcm\right)$

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TM

Trần Minh Anh

15 tháng 1 2017

cảm ơn bn nha ^.^!!!

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KN

kiwi nguyễn

26 tháng 6 2019

CMR:

a) $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

b) Cho A = $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

CMR: $\frac{7}{12}< A< \frac{5}{6}$

#Toán lớp 7

1

S

svtkvtm

26 tháng 6 2019

$\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)$

$theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)$

$A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)$ $\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}$

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AQ

Alex Queeny

14 tháng 9 2015 - olm

Chứng minh $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

0

IM

I miss my love

25 tháng 11 2017 - olm

chứng minh rằng $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$$\frac{1}{50}$

#Toán lớp 7

1

NQ

Nguyễn Quang Đức

25 tháng 11 2017

Ta có: $\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}$

$-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}$

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LH

Lê Hà Phương

9 tháng 9 2015 - olm

Chứng minh: $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

1

SN

✓ ℍɠŞ_ŦƦùM $₦G ✓

9 tháng 9 2015

$\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}$

$=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+...+\frac{1}{50}$

=>đpcm

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NM

Nguyễn Minh Tuấn

21 tháng 10 2015 - olm

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..........+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+..........+\frac{1}{50}$

Chứng minh

#Toán lớp 7

1

M

Miyuhara

21 tháng 10 2015

$\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)$

$=\left(1+\frac{1}{2}+..+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}$

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TN

Thanh Nguyen

30 tháng 11 2015 - olm

chứng minh rằng $\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}$

#Toán lớp 7

1

NH

Nguyễn Hưng Phát

30 tháng 11 2015

Mới nhìn mà muốn vào viện ngay rồi

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