Các câu hỏi tương tự
( 1 / 4 . 9 + 1 / 9 . 14 + 1 / 14 . 19 + ... + 1 / 44 .49 ) . (1 - 3 - 5 - 7 - ... - 49) / 89 = ?
TÍNH A=[(1/4*9)+(1/9*14)+(1/14*19)+...+(1/44*49)] x (1-3-5-7-...-49)/89
(1/4*9+1/9*14+...+1/44*49)*(1-3-5-7-...-49/89)
\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+\frac{1}{19\cdot24}+.....+\frac{1}{44\cdot49}\right)1\frac{-3-5-7-...-49}{89}\)
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
\(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
Tính:
\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
Bài 1: tính
1, 1-5+7-8+4-1+5-7+8
2, 14-23+(5+14)-(5-23)+17
3, 12-44+9-3+14-19-9-12
4,22-(4-8+12)+(-8-12+4)
Chứng tỏ rằng:
\(M=\left(\frac{1}{4}+\frac{1}{9.4}+\frac{1}{14.49}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-9-...-49}{89}=-\frac{9}{28}\)