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\(A=2^4+4^4+6^4+...+18^4+20^4\)
\(=2^4\left(1^4+2^4+3^4+...+9^4+10^4\right)\)
\(=16.25333=405328\)
\(2^2+4^2+6^2+...+20^2\)
=\(\left(1.2\right)^4+\left(2.2\right)^4+\left(3.2\right)^4+...+\left(2.10\right)^4\)
=\(1^4.2^4+2^4.2^4+3^4.2^{\text{4}}+....+10^4.2^4\)
=\(2^4.\left(1^4+2^4+3^4+...+10^4\right)\)
=16.25333=405328
S = 2^4.(1^4+2^4+3^4+.....+10^4)
= 16 . 25333
= 405328
Tk mk nha
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: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
Chuyển 14 thành 1. Và coi ... là x. Vậy:
\(1^4+1^4+x+9^4+10^4=\)\(25333\)
\(1+1+x+9^4+10^4=25333\)
\(\left(1+1\right)+x+9^4+10^4=\)\(25333\)
\(2+x+6561+10000=\)\(25333\)
\(x+\left(2+6561+10000\right)=\)\(25333\)
\(x+16563=25333\)
\(x=25333-16563\)
\(x=8770\)
Vậy số cần điền vào chỗ trống là 8770.
Học tốt nha.
cảm ơn bạn .ai đây ạ ........................................................................................................................................................................................ạ..........................