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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
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
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..........................