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Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)\)
\(A=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
A = 1.2 + 2.3 + .... + 999.1000
3A = 1.2.3 + 2.3.(4-1) + .... + 999.1000.(1001 - 998)
3A = 1.2.3 + 2.3.4 - 1.2,3 +..... + 999.1000.1001 - 998.999.1000
3A = 999 . 1000 . 1001
A = 333 x 1000 x 1001 = 333 333 000
3A=1x2x3+2x3x(4-1)+3x4x(5-2)+.............+999x1000x(1001-998)
3A=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+............+999x1000x1001-998x999x1000
3A=999x1000x1001
A=999x1000x1001:3
A=333333000
\(B=1.2+2.3+3.4+...+18.19\)
\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+18.19.\left(20-17\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+18.19.20-17.18.19\)
\(3B=18.19.20\Rightarrow B=\frac{18.19.20}{3}=2280\)
Ta có : B = 1 x 2 + 2 x 3 + 3 x 4 +.... + 18 x 19
=> 3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 18 x 19 x 3
3B = 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 18 x 19 x (20 - 17)
3B = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x3 + 3 x 4 x5 - 2 x 3 x 4 + ... + 18 x 19 x 20 - 17 x 18 x 19
3B = 18 x 19 x 20
B = 18 x 19 x 20 : 3
= 2280
Vậy B = 2280