Đáp án:
Giải thích các bước giải:
9/ \(A=\dfrac{3^2}{10}+\dfrac{3^2}{40}+\dfrac{3^2}{88}+...+\dfrac{3^2}{340}\)
⇒ \(A=3\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}\right)\)
⇒ \(A=3.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
⇒ \(A=3.\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
⇒ \(A=\dfrac{27}{20}\)
10/
\(A=\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{26\cdot31}\)
\(A=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\right)\)
\(A=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(A=5\left(1-\dfrac{1}{31}\right)=5\cdot\dfrac{30}{31}=\dfrac{150}{31}\)
11/
$A =\dfrac{6}{15} + \dfrac{6}{35} + \dfrac{6}{63} + \dfrac{6}{99} + \dfrac{6}{143}$
$A = 3.(\dfrac{2}{3.5} + \dfrac{2}{5.7} + \dfrac{2}{7.9} + \frac{2}{9.11} + \dfrac{2}{11.13})$
$A = 3.(\dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{9} + \dfrac{1}{9} - \dfrac{1}{11} + \dfrac{1}{11} - \dfrac{1}{13})$
$A = 3.(\dfrac{1}{3} - \dfrac{1}{13})$
$A = 1 - \dfrac{3}{13}$
$A = \dfrac{13}{13} - \dfrac{3}{13}$
$A = \dfrac{10}{13}$
12/
A = \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{49.51}\)
2A = \(\dfrac{3.2}{3.5}+\dfrac{3.2}{5.7}+\dfrac{3.2}{7.9}+...+\dfrac{3.2}{49.51}\)
2A = 3\(\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{49.51}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
2A = 3 . \(\dfrac{16}{51}\)
2A = \(\dfrac{16}{17}\)
=> A = \(\dfrac{16}{17}\) : 2
=> A = \(\dfrac{8}{17}\)
14/
$A=\dfrac{1}{2}$+ $\dfrac{2}{8}$+ $\dfrac{3}{28}$+$\dfrac{4}{77}$+ $\dfrac{5}{176}$+$\dfrac{6}{352}$
A = $\dfrac{1}{2}$+$\dfrac{1}{4}$+$\dfrac{3}{28}$+$\dfrac{4}{77}$+ $\dfrac{5}{176}$+$\dfrac{3}{176}$
A = $\dfrac{3}{4}$ + $\dfrac{7}{44}$+$\dfrac{1}{22}$
A = $\dfrac{21}{22}$
