biết \(1^2=2^2+3^2+...10^2=385\) Hãy tính:
A= \(2^2+4^2+6^2+..+20^2\)và B = \(1^2+3^2+6^2+9^2+...+30^2\)
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a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
A = \(2^2.\left(1^2+2^2+3^2+...+10^2\right)=4.385=1540\)
B=\(3^2.\left(1^2+2^2+3^2+...+10^2\right)=385.9=3465\)
\(S=2^2+4^2+....+20^2=?\)
\(=\left(2.1\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+....+\left(2.10\right)^2\)
\(=2^2.1^2+2^2.2^2+2^2.2^3+...+2^2.10^2\)
\(=2^2.\left(1^2+2^2+3^2+...+10^2\right)\)
\(=2^2.385\)
\(=4.385\)
\(=1540\)
S=22+42+...+202
=> 1/2 .S=12+22+...+102
=> 1/2 .S=385
=> S = 385 . 2
=> S = 770
Lời giải:
$S=10^2+(10.2)^2+(10.3)^2+...+(10.9)^2+(10.10)^2$
$=10^2(1^2+2^2+3^2+...+9^2+10^2)$
$=100.385=38500$
\(P=3^2+6^2+9^2+...+30^2\)
\(=\left(1.3\right)^2+\left(2.3\right)^2+\left(3.3\right)^2+...+\left(10.3\right)^2\)
\(=\left(1^2+2^2+3^2+...+10^2\right).3^2\)
\(=385.9\)
\(=3465\)
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: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
Ta có : 12+22+32+...+92+102=385
22+42+62+...+202=22.12+22.22+22.32+....+22.92+22.102
=22.(11+22+32+....+92+102)
=4.385
=1540
S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385 = 1540
A = 3² + 6² + 9² + ... + 30²
= (3.1)² + (3.2)² + (3.3)² + ... + (3.10)²
= 3².(1² + 2² + 3² + ... + 10²)
= 9.385
= 3465
ai nhanh nhất,đúng nhất mình cho