K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

22 tháng 8 2023

Ta có: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989A=(491+9141+14191+...+44491)891357...49

⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989A=51(495+9145+14195+...+44495)891357...49

⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989A=51(4191+91141+141191+...+441491)891357...49

⇔�=15⋅(14−149)⋅1−3−5−7−...−4989A=51(41491)891357...49

⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989A=51(449494)891357...49

⇔�=15⋅45196⋅1−3−5−7−...−4989A=5119645891357...49

⇔�=9196⋅1−3−5−7−...−4989A=1969891357...49

⇔�=9196⋅−62389=−928A=196989623=289
 

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

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

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

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

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

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

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

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

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

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 \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)

20 tháng 11 2018

\(=\left[\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right)+\frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right)+\frac{1}{5}\left(\frac{1}{14}-\frac{1}{19}\right)+...+\frac{1}{5}\left(\frac{1}{44}-\frac{1}{49}\right)\right]\cdot\frac{1-\left(3+5+...+49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(52+52+...+52\right)\left\{12\text{ số 52}\right\}}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)

\(=\frac{9}{196}\cdot-7=\frac{9}{28}\)

8 tháng 11 2015

\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(49+3\right)\left(\left(49-3\right):2+1\right):2}{89}\)

\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-26.24}{89}=\frac{45}{4.5.49}.\frac{-623}{89}=-\frac{9}{28}\)

8 tháng 11 2015

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

\(\Rightarrow5A=5.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)

\(=\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right).\frac{1+\frac{\left(-3-47\right).23}{2}-49}{89}\)

\(=\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1+\left(-575\right)-49}{89}\)

\(=\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}=\frac{45}{196}.\left(-7\right)=-\frac{45}{26}\)

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)=\(\dfrac{-9}{28}\)

29 tháng 3 2017

fty

22 tháng 8 2023

\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)

22 tháng 8 2023

Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5

 

29 tháng 6 2018

câu B là \(2^{12}\) nha mấy bn

30 tháng 6 2018

Bài 1:

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

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

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).-\left(\frac{\left(49+3\right).24:2-1}{89}\right)\)(Do tổng có 24 số)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).-\left(\frac{52.12-1}{89}\right)\)

\(=\frac{1}{5}.\frac{45}{196}.\left(-7\right)=-\frac{9}{28}\)

Bài 2:

a) Ta có:

\(|2x+3|=x+2\)

<=> x + 2 >=0 và: \(\orbr{\begin{cases}2x+3=x+2\\2x+3=-x-2\end{cases}}\)

<=> x >= -2 và \(\orbr{\begin{cases}2x-x=2-3\\2x+x=-2-3\end{cases}}\)

<=> x >= -2 và \(\orbr{\begin{cases}x=-1\left(n\right)\\x=-\frac{5}{3}\left(n\right)\end{cases}}\)( n là viết tắt của "nhận" nha bạn)

Vậy x ={-1 ; -5/3}

Xin lỗi vì tớ ko thể lồng dấu \(\hept{\begin{cases}\\\end{cases}}\) và dấu \(\orbr{\begin{cases}\\\end{cases}}\) được nếu lồng sẽ bị lỗi nên tớ dùng chữ "và" nha bạn

b) 

A = \(|x-2006|+|2007-x|\)

Vì \(\hept{\begin{cases}|x-2006|\ge0\\|2007-x|\ge0\end{cases}}\)

Nến giá trị A sẽ nhỏ nhất khi \(\orbr{\begin{cases}x=2006\\x=2007\end{cases}}\)

=> Min A = 1 khi x ={2006 ; 2007}