\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+....+\dfrac{3}{59.61}\)

\(S=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{59}-\dfrac{1}{61}\)

\(S=\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(S=\dfrac{1}{5}-\dfrac{1}{61}\)

\(S=\dfrac{56}{305}\)

Vậy S = \(\dfrac{56}{305}\)

\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)

\(S=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(S=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}.\dfrac{56}{305}=\dfrac{84}{305}\)

S = \(\dfrac{3}{1.2}\)+\(\dfrac{3}{2.3}\)+\(\dfrac{3}{3.4}\)+\(\dfrac{3}{4.5}\)+...+\(\dfrac{3}{2015.2016}\)

= 3.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\right)\)

= 3.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

= 3.\(\left(1-\dfrac{1}{2016}\right)\) = 3.\(\dfrac{2015}{2016}\)=\(\dfrac{3.2015}{2016}\)=\(\dfrac{1.2015}{672}\)=\(\dfrac{2015}{672}\)

Vậy S = \(\dfrac{2015}{672}\)

Ta có S=\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+\dfrac{3}{4.5}+...+\dfrac{3}{2015.2016}\)

=3.(\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\))

=3.(\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\))

=\(3.\left(1-\dfrac{1}{2016}\right)\)

= \(3-\dfrac{1}{672}\)=\(\dfrac{2015}{672}=2\dfrac{671}{672}\)

\(M=\dfrac{5^3}{1\cdot6}+\dfrac{5^3}{6\cdot11}+...+\dfrac{5^3}{26\cdot31}\)

\(=5^2\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)

\(=5^2\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5^2\left(1-\dfrac{1}{31}\right)\)\(=25\cdot\dfrac{30}{31}=\dfrac{750}{31}\)

3/ Chu vi hình chữ nhật:

\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)

Diện tích hình chữ nhật:

\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)

Đơn vị trong ngoặc ghi là đơn vị diện tích nhá!

a, Ta có: \(\dfrac{32}{37}>\dfrac{32}{54}>\dfrac{19}{54}\Rightarrow\dfrac{32}{37}>\dfrac{19}{54}\)

b, Ta có: \(\dfrac{18}{53}>\dfrac{18}{54}=\dfrac{1}{3}\Rightarrow\dfrac{18}{53}>\dfrac{1}{3}\left(1\right)\)

\(\dfrac{26}{78}=\dfrac{1}{3}\left(2\right)\)

Từ (1) và (2) ta suy ra \(\dfrac{18}{53}>\dfrac{26}{78}\)

c, Ta thấy: \(\dfrac{25}{103}< \dfrac{25}{100}=\dfrac{1}{4}\left(1\right)\)

\(\dfrac{74}{295}>\dfrac{74}{296}=\dfrac{1}{4}\left(2\right)\)

Từ (1) và (2) ta suy ra \(\dfrac{25}{103}< \dfrac{74}{295}\)

tick cho mk với

Ta có:A-1=\(\dfrac{10^8+2}{10^8-1}-1=\dfrac{10^8+2-10^8+1}{10^8-1}=\dfrac{3}{10^8-1}\)

B-1=\(\dfrac{10^8}{10^8-3}-1=\dfrac{10^8-10^8+3}{10^8-3}=\dfrac{3}{10^8-3}\)

Do \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\)

=>A-1>B-1

<=>A>B

Vậy...

mik cũng đg cần mà bnXuân Tuấn Trịnh làm đúng ko z

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

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

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

A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-...-\left(\dfrac{1}{65}-\dfrac{1}{65}\right)-\dfrac{1}{68}\right]\)

A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{68}\right]\)

A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\dfrac{1}{68}\right]\)

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

A = \(\dfrac{11}{17}\)

1/3.(1/2.5+1.5.8+1/8.11+...+1/65.68)

=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/65-1/68)

=1/3(1/2-1/68)

=1/3.33/68

=11/68

nhớ theo dõi mik nha