K
Khách

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

24 tháng 8 2023

Sửa đề:

\(A=\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\)

\(A=2.\left(\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\right)\)

\(A=\dfrac{6}{10}+\dfrac{6}{40}+\dfrac{6}{88}+\dfrac{6}{154}\)

\(A=6.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)\)

\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)\)

\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\)

\(A=6.\dfrac{6}{14}\)

\(A=\dfrac{36}{14}=\dfrac{18}{7}\)

\(=\dfrac{1}{5}.3+\dfrac{1}{5}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{7}\)

\(=\dfrac{1}{5}.\left(3+\dfrac{3}{4}\right)+\dfrac{1}{11}.\left(\dfrac{3}{4}+\dfrac{3}{7}\right)\)

\(=\dfrac{1}{5}.\dfrac{15}{4}+\dfrac{1}{11}.\dfrac{33}{28}=\dfrac{3}{4}+\dfrac{3}{28}=\dfrac{6}{7}\)

25 tháng 8 2023

Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

25 tháng 8 2023

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)\(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)\(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)\(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\)\(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)

25 tháng 8 2023

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

25 tháng 8 2023

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

20 tháng 8 2023

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

AH
Akai Haruma
Giáo viên
22 tháng 8 2023

Lời giải:
Hiệu đáy lớn và đáy bé là:
$141\times 2: 23=\frac{282}{23}$ (m) 

Đáy bé hình thang:

$\frac{282}{23}: (5-3).3=\frac{423}{23}$ (m)

Đáy lớn hình thang:
$\frac{282}{23}: (5-3).5=\frac{705}{23}$ (m)

Diện tích hình thang lúc đầu:

$(\frac{423}{23}+\frac{705}{23}).23:2=564$ (m2)

20 tháng 11 2017

1.

\(B=20182018.2017-20172017.2018\)

\(B=2018.10001.2017-2017.10001.2018\)

\(B=0\)

20 tháng 11 2017

1 B=20182018.2017-20172017.2018

B=2018.10001.2017-2017.10001.2018

B=0

2 C=12+22+32+...+1002

C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)

C=1+2+1.2+3+2.3+...+100+99.100

C=(1+2+3+...+100)+(1.2+2.3+...+99.100)

C=[(1+100).100:2]+[(99.100.101):3]

C=5050+333300

C=338350

HQ
Hà Quang Minh
Giáo viên
8 tháng 10 2023

\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)

1 tháng 5 2021

-3/11.(-22)/66.121/15

=(-3).(-22).121

 11.66.15

=11

15

3/7.2/5.7/3.20.19/72

=3.2.7.20.19

7.5.3.72

=76

16

6/7.8/13+6/13.9/7-3/13.6/7

=6/7.8/13+6/7.9/13-3/13.6/7

=6/7.(8/13+9/13-3/13)

=6/7.14/13

=12/13

-1/4.152/11+68/4.(-1)/11

=152/4.(-1)/11+68/4.(-1)/11

=(-1)/11.(152/4+68/4)

=(-1)/11.220/4

=-110/22

-5/7.2/11+(-5)/7.9/11+12/7

=-5/7.2/11+-5/7.9/11+12/7

=-5/7.(2/11+9/11)+12/7

=-5/7.1+12/7

=(-5)/7+12/7

=7/7

=1

146/13-(18/7+68/13)

=146/13-18/7-68/13

=(146/13-68/13)-18/7

=78/13-18/7

=6-18/7

=42/7-18/7

=24/7

\(25-\left(44-756+12\right)+\left(44-756+12\right)\)

\(=25-44+756-12+44-756+12\)

\(=25+\left(44-44\right)+\left(756-756\right)+\left(12-12\right)\)

=25