Question

Tính giá trị của các biểu thức sau:

a) A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{299.300}\)

b) B = \(\dfrac{10^2}{16.26}+\dfrac{10^2}{26.36}+\dfrac{10^2}{36.46}+...+\dfrac{10^2}{86.96}\)

Answer 1

a, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{299.300}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

\(=1-\dfrac{1}{300}=\dfrac{299}{300}\)

Vậy \(A=\dfrac{299}{300}\)

b, \(B=\dfrac{10^2}{16.26}+\dfrac{10^2}{26.36}+...+\dfrac{10^2}{86.96}\)

\(=10\left(\dfrac{10}{16.26}+\dfrac{10}{26.36}+...+\dfrac{10}{86.96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+...+\dfrac{1}{86}-\dfrac{1}{96}\right)\)

\(=10\left(\dfrac{1}{16}-\dfrac{1}{96}\right)\)

\(=10.\dfrac{5}{96}=\dfrac{25}{48}\)

Vậy...

Answer 2

a,\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{299.300}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(A=\dfrac{1}{1}-\dfrac{1}{300}=\dfrac{299}{300}\)

Answer 3

a) \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3.4}+....+\dfrac{1}{299\cdot300}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{299}-\dfrac{1}{300}\)

\(=1-\dfrac{1}{300}=\dfrac{299}{300}\)

b) \(B=\dfrac{10^2}{16\cdot26}+\dfrac{10^2}{26\cdot36}+\dfrac{10^2}{36\cdot46}+...+\dfrac{10^2}{86.96}\)

\(\dfrac{B}{10}=\dfrac{10}{16\cdot26}+\dfrac{10}{26\cdot36}+\dfrac{10}{36\cdot46}+...+\dfrac{10}{86\cdot96}\)

\(\dfrac{B}{10}=\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+\dfrac{1}{36}-\dfrac{1}{46}+...+\dfrac{1}{86}-\dfrac{1}{96}\)

\(\dfrac{B}{10}=\dfrac{1}{16}-\dfrac{1}{96}=\dfrac{5}{96}\)

\(\Rightarrow B=\dfrac{5}{96}\cdot10=\dfrac{25}{48}\)

Answer 4

b, \(B=\dfrac{10^2}{16.26}+\dfrac{10^2}{26.36}+......+\dfrac{10^2}{86.96}\)

\(B=10.\left(\dfrac{10}{16.26}+\dfrac{10}{26.36}+......+\dfrac{10}{86.96}\right)\)

\(B=10.\left(\dfrac{1}{16}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{36}+......+\dfrac{1}{86}-\dfrac{1}{96}\right)\)

( do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)

\(B=10.\left(\dfrac{1}{16}-\dfrac{1}{96}\right)=10.\dfrac{5}{96}=\dfrac{25}{48}\)

Chúc bạn học tốt!!!