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Ta thấy A gấp 12+22+....+102 4 lần nên Tổng A gấp 4 lần nó
=> A=385.4=1540
Đặt S = \(1^2+2^2+3^2+\cdots+\) \(10^2\)
Ta có: \(2^2=(2\cdot1)^2=\) \(4\cdot1^2\)
\(4^2=(2\cdot2)^2\) \(=4\cdot2^2\)
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\(20^2=(2\cdot10)^2=4\cdot10^2\)
=> \(A=4\cdot1^2+4\cdot2^2+4\cdot3^2+\cdots+4\cdot10^2\)
\(A=4\cdot\left(1^2+2^2+3^2+\cdots+10^2\right)=4S\)
\(\Rightarrow4S=4\cdot385\)
\(4S=1540\)
\(\Rightarrow A=1540\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
\(A=\left(12^2+14^2+16^2+18^2+20^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\)
\(A=\left(12^2-1^2\right)+\left(14^2-3^2\right)+\left(16^2-5^2\right)+\left(18^2-7^2\right)+\left(20^2-9^2\right)\)\(A=\left(12+1\right)\left(12-1\right)+\left(14+3\right)\left(14-3\right)+\left(16-5\right)\left(16+5\right)+\left(18-7\right)\left(18+7\right)+\left(20-9\right)\left(20+9\right)\)
\(A=11.13+11.17+11.21+11.25+11.29\)
\(A=11.\left(13+17+21+25+29\right)\)
\(A=11.\left[\left(13+17\right)+\left(21+29\right)+25\right]\)
\(A=11.\left(30+50+25\right)\)
\(A=11.105=1155\)
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