C=\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}.............+\frac{1}{2017x2018}\)
Tính nhanh:
\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
Tính:
\(A=\frac{1x2+2x3+3x4+...+2016x2017}{2017x2018}\)
Gọi B = 1x2 + 2 x 3 + 3 x 4 + ... + 2016 x2017
3B = 3 x ( 1x2 + 2x3 + 3x4 + ... + 2016x2017)
= 1x2x3 + 2x3x3 + 3x4x3 + ... + 2016x2017x3 )
= 1x2x3 + 2x3x( 4-1) + 3x4x( 5 -2 ) + ... + 2016x2017x( 2018 - 2015)
= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 2016x2017x2018 - 2015x2016x2017
= 2016 x2017 x2018
B = 672 x2017 x2018
Mà A = \(\frac{672x2017x2018}{2017x2018}\)
= 672
Vậy A = 672
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+..............+\frac{1}{8x9}=?\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
=\(1-\frac{1}{9}\)
=\(\frac{8}{9}\)
OK XONG NHỚ CHO MIK NHA
\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)
=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
=1-\(\frac{1}{9}\)
=\(\frac{8}{9}\)
\(\frac{1}{1\times2}+........+\frac{1}{8\times9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{99x100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
k cho mình nha bạn
=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
A=\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}\)
A = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}\)
A = \(\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)
Mk đầu tiên nha
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{999x1000}+1\)\(1=?\)
minh ko biet xin loi bn nha!
minh ko biet xin loi bn nha!
minh ko biet xin loi bn nha!
minh ko biet xin loi bn nha!
tính nhanh :\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{7x8}+\frac{1}{8x9}\)
Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)
bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2
tính nhanh
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+....+\frac{1}{99x100}\)
1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\)