A=$\dfrac{1}{4.9} \dfrac{1}{9.14} \dfrac{1}{14.19} ... \dfrac{1}{44.49} \left(\dfrac{1-3-5-7-...-49}{89}\right)$ B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6 8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^...

TL

Trần Lê Nhi

29 tháng 6 2018

...

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#Toán lớp 7

1

TL

Trần Lê Nhi

29 tháng 6 2018

câu B là $2^{12}$ nha mấy bn

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OI

✞ঔৣ۝????à ????????ị????۝ঔৣ✞

10 tháng 2 2021

A=$\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\dfrac{1-3-5-7-...-49}{89}$

#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 2 2021

Ta có: $A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}$

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DH

❆❆❉→Đào Hoàng Lâm Tuyền←❆❆❆

30 tháng 1 2022

$\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\dfrac{1-3-5-7-...-49}{89}$

thực hiện phép tính

#Toán lớp 7

2

NT

Nguyễn Thái Thịnh

30 tháng 1 2022

Đặt $A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}$

$=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}$

$=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}$

$=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}$

$=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}$

Đặt $B=1-3-5-7-..-49$

$=1-\left(3+5+7+...+49\right)$

$=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}$

$=1-624$

$=-623$

$\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}$

Vậy: $\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}$

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TD

Trần Đức Huy

30 tháng 1 2022

Xét $\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)$

=$\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)$

=$\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)$

=$\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)$

=$\dfrac{1}{5}.\dfrac{45}{196}$

=$\dfrac{9}{196}$

Xét $\dfrac{1-3-5-7-..-49}{89}$

=$\dfrac{1-\left(3+5+7+...+49\right)}{89}$

CT tính sl số hạng (số cuối - số đầu ):2+1

số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24

Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2

=> tổng = [(3+49).24]:2=624

=>$\dfrac{1-624}{89}$

=$\dfrac{-623}{89}$

=-7

từ đó ta có $\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}$

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AS

★彡 A͛r͛a͛k͛i͛ S͛e͛i͛j͛u͛r͛o͛ 彡★

11 tháng 2 2018

Thực hiện phép tính:

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$B=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\dfrac{2^2}{7^2}-\dfrac{4}{343}}$

#Toán lớp 7

3

D

Despacito

11 tháng 2 2018

$B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}$

$B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}$

$B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}$

$B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}$

$B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}$

$B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}$

$B=\frac{300}{343}:\frac{1347}{343}$

$B=\frac{100}{449}$

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M

Mafia

11 tháng 2 2018

$A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}$

$A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}$

$A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}$

$A=\frac{-1}{2}+\frac{1710}{9}$

$A=\frac{-1}{2}+190$

$A=\frac{-1}{2}+\frac{380}{2}$

$A=\frac{379}{2}$

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TN

Tú Nguyễn Văn

11 tháng 2 2018

Thực hiện phép tính:

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$B=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\dfrac{2^2}{7^2}-\dfrac{4}{343}}$

#Toán lớp 7

1

GH

Gia Hân Ngô

11 tháng 2 2018

$A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$

$=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}$

$=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}$

= $\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}$

= $\dfrac{2}{3.4}-\dfrac{5\left(-6\right)}{9}$

= $\dfrac{7}{2}$

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TC

Ta Chia Tay Đi

9 tháng 10 2017

Tính :

c) C = $\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}$

d) D = $\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\left(\dfrac{1-3-5-7-...-49}{89}\right)$

#Toán lớp 7

0

TT

Trần Thị Hảo

2 tháng 8 2017

Bài 1 : Rút gọn$M=\dfrac{1^{2016}+2^{2016}+3^{2016}+...+10^{2016}}{2^{2016}+4^{2016}+6^{2016}+...+20^{2016}}$Bài 2 : Tính$N=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-9}{89}$$P=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6-8^4.3^5}-\dfrac{5^{10}.7^3-25^5.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}$Bài 3: Rút gọna) A = |2x + 4,6| - 2x + 15,4b) B = |x + 7,2| - |x - 1,2|c) C = 8,5x - 19, 5 - |1,5x + 4,5|d) D = 8,5 + x...

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#Toán lớp 7

0

VN

Võ Ngọc Phương

VIP

22 tháng 8 2023

Tính:

$A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}$

Giải chi tiết dùm mik nha. Thankss

#Toán lớp 7

4

NM

Nguyễn Minh Dương

VIP

22 tháng 8 2023

$A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)$

$A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)$

Ta đặt: $P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.$

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NN

Nguyễn Nhân Dương

22 tháng 8 2023

Ta có: $A = (4 \cdot 9 1 + 9 \cdot 14 1 + 14 \cdot 19 1 + ... + 44 \cdot 49 1) \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 5 1 \cdot (4 \cdot 9 5 + 9 \cdot 14 5 + 14 \cdot 19 5 + ... + 44 \cdot 49 5) \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 5 1 \cdot (4 1 - 9 1 + 9 1 - 14 1 + 14 1 - 19 1 + ... + 44 1 - 49 1) \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 5 1 \cdot (4 1 - 49 1) \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 5 1 \cdot (4 \cdot 49 49 - 4) \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 5 1 \cdot 196 45 \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 196 9 \cdot 89 1 - 3 - 5 - 7 - ... - 49$

$\Leftrightarrow A = 196 9 \cdot 89 - 623 = - 28 9$

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DT

Duong Thi Nhuong

8 tháng 3 2017

a) Tìm $n\in N$* :

+) $\dfrac{1}{8}.16^n=2^n$

+) $27< 3^n< 243$

b) Tính : $\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}$

#Toán lớp 8

1

DM

Đức Minh

8 tháng 3 2017

Bài 1 :

a) +) $\dfrac{1}{8}\cdot16^n=2^n$

$\Leftrightarrow\dfrac{1}{8}=\dfrac{2^n}{16^n}$

$\Rightarrow\dfrac{1}{8}=\dfrac{1}{8}^n$

Vậy n = 1.

+) $27< 3^n< 243$

$\Leftrightarrow3^3< 3^n< 3^5$

Vậy n = 4.

Bài 2 : $\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}$

$\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}$

$\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{-623}{89}=-\dfrac{45}{28}$

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N

ngonhuminh

10 tháng 3 2017

Bài 2 :

chưa hiểu: @Duc Minh

$\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)=\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}+...-\dfrac{1}{49}\right)$

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LT

Lê Thị Thùy Dung

28 tháng 3 2017

$\left(\dfrac{1}{4}.9+\dfrac{1}{9.14}+\dfrac{1}{14.19}+....+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}$

GIÚP với các bn!

#Toán lớp 7

2

CP

Chi pipi

28 tháng 3 2017

=$\dfrac{1}{5}$.($\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+....+\dfrac{5}{44.49}$).$\dfrac{1-\left(3+5+7+...+49\right)}{89}$

=$\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)$.$\dfrac{1-624}{89}$

=$\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right)$.(-7)

=$\dfrac{1}{5}$.$\dfrac{45}{196}$.(-7)=$\dfrac{-9}{28}$

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LT

Lê Thị Thùy Dung

29 tháng 3 2017

fty

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