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A = 1 x 2 + 2 x 3 + ....... + 10 x 11
3A = 1 x 2 x 3 + 2 x 3 x 3 + ..........+ 10 x 11 x 3
3A = 1 x 2 x (3-0) + 2 x 3 x (4-1) + .......... + 10 x 11 x (12 -9)
3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ........... + 10 x 11 x 12 - 9 x 10 x 11
3A = (1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) +............ + 10 x 11 x 12
3A = 10 x 11 x 12 = 1320
A = 1320 : 3 = 440
Gọi biểu thức trên là A, ta có :
A= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
1/5x6+1/6x7+1/7x8+...+1/2019x2020
= 6/5+7/6+8/7+...+2020/2019
Rút gọn cho nhau ta còn 2020/5=404
1/5 x 6 + 1/6 x7 + 1/7 x8 + ... + 1/2019 - 1/ 2020
=1/5 -1/6 +1/6 -1/7 + 1/7 - 1/8 + ... + 1/2019 - 1/2020
Sau khi giản ước, ta còn:
1/5 - 1/2020 = 403/2020.
Đáp số: 403/2020
1/5×6 + 1/6×7 + 1/7×8 + ... + 1/24×25
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 5/25 - 1/25
= 4/25
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
đặt A=1x2 + 3x4 + 5x6 + 7x8 + 9x10 + 11x12
A=1x(3-1)+3x(5-1)+...+9x(11-1)+11x(13-1)
=1x3-1+3x5-3+...+9x11-9+11x13-11
=(1x3+3x5+...9x11+11x13)-(1+3+5+...+11)
đặt C=1x3+3x5+...+9x11+11x13
6C=6x(1x3+3x5+...+9x11+11x13)
=1x3x6+3x5x(7-1)+...+9x11x(13-7)+11x13x(15-9)
=1x3x6+3x5x7-1x3x5+...+9x11x13-7x9x11+11x13x15-9x11x13
loại các số giống nhau ta được:
=11x13x15+(1x3x6-1x3x5)
=11x13x15+1x3x(6-5)
=11x13x15+1x3x1
=11x13x15+3
Vậy C=2538:6=423
=>A=423+1+3+5+...+11
=423+36
Vậy A=459
1x2 + 3x4 + 5x6 + 7x8 + 9x10
=2 + 12 +30 +56 +90
=14 + 30 + 56 + 90
=44 +56 + 90
=100 + 90
=190
tk nhé
= 1x2 + 3x2x2 + 5x3x2 + 7x4x2 + 9x5x2 + 11x6x2 + 13x7x2 + 15x8x2
=1x2 + 6x2 + 15 x2 + 28x2 + 45x2 + 66x2 + 91x2 + 120x2
=2 x ( 1+6+15+28+45+66+91+120)
= 2 x 370
=740
=
Dặt \(A=\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{6}-\frac{1}{2020}\)
\(A=\frac{1007}{6060}\)
hok tốt!!