Sửa đề:

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

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

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

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

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

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

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 = \(\dfrac{11}{17}\)

bạn kiểm tra lại đề nhé.

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

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

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

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

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

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

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

Sửa đề:

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

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

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

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

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

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

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

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

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

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

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

kết quả của mình là:16493,1

kết quả của bài trên là:

243,45x800x2=16578

đáp số:16578

\(=\frac{2014}{2015}\)

tk mk nha

Ta có : 

4/1 . 5 + 4/5 . 9 + ...+ 4/2001 . 2005 

= 1 - 1/5 + 1/5 - 1/9 + ...+ 1/2001 - 1/2005 

= 1 - 1/2005

=  2004/2005 

Tham khảo nha !!! 

lộn

\(=\frac{2004}{2005}\)

xl nha

Ta có : A = 1 + 4 + 4² + 4³ + ..... + 4²³

=> 4A = 4 + 4² + 4³ + ..... + 4²⁴ 

=> 4A - A = 4²⁴ - 1

=> 3A = 4²⁴ - 1

=> 3A + 1 = 4²⁴ = (4³)⁸ = 64⁸ > 63⁷

Vậy 3A + 1 > 63⁷

Ta co A =4^0+4^1+...+4^23

lai co 4A=4(4^0+4^1+4^2+...+4^23)

         4A=4^1+4^2+...4^24

Mà 3A=4A-A=(4^1+4^2+...4^24)-(4^0+4^1+...+4^23)

3A=4^24-4^0=4^24-1

3A+1=4^24-1+1+4^24

khúc sau đổi về rồi so sánh 

nhớ nhá

4A=4¹+4²+4³+4⁴+...+4²⁴

4A-A=3A=(4¹+4²+4³+4⁴+...+4^24)_(4⁰+4¹+4²+4³+...+4²³)=(4¹-4¹)+(4²-4²)+...+(4²³-4²³)+(4²⁴-4⁰)=4²⁴-1

3A+1=4²⁴-1+1=4²⁴=4^3.8=(4³)⁸=64⁸>64⁷

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)