Question

a ( 1+2+3+....+100) * (1^2+ 2^2 + 3^3+....+100^2) * ( 65. 111- 13. 15. 37)

b  1999. 1999. 1998- 1998. 1998. 1999

Answer 1

\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(65.111-13.15.37\right)\)

\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).\left(7215-7215\right)\)

\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^3+...+100^2\right).0\)

\(=0\)

\(1999.1999.1998-1998.1998.1999\)

\(=1999.1998.\left(1999-1998\right)\)

\(=1999.1998.1\)

Tham khảo nhé~

Answer 2

13453 nhe

Answer 3

bn oi viet cach lm ra giup mk voi