Tính: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+19}=?\)
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21 tháng 4 2017
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
12 tháng 5 2019
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
24 tháng 8 2019
Ta có phần tử \(=\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\)
\(=\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+\left(\frac{19}{1}+1\right)-19\)
\(=\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{1}+\frac{20}{20}-20\)
\(=20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\left(1\right)\)
Thay (1) vào P ta được :
\(P=\frac{20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}}\)
\(=20\)
D
28 tháng 4 2015
Tử số = T = \(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+....+\frac{18}{2}+\frac{19}{1}\)
\(=\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+\left(\frac{3}{17}+1\right)+....+\left(\frac{19}{1}+1\right)-19\)
\(=\frac{20}{19}+\frac{20}{18}+\frac{20}{17}+....+\frac{20}{2}+20-19\)
\(=\frac{20}{2}+\frac{20}{3}+....+\frac{20}{18}+\frac{20}{19}+\frac{20}{20}\)
\(=20\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
= 20.Mẫu số
\(\Rightarrow\frac{\frac{1}{19}+\frac{2}{18}+....+\frac{18}{2}+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{19}+\frac{1}{20}}=20\)
23 tháng 5 2016
Mẫu số = 1/19 + 2/18 + 3/17 + ... + 18/2 + 19/1
= ( 1/19 + 2/18 + 3/17 + ... + 18/2 ) + ( 1 + 1 + ... + 1 )
( 18 phân số ) ( 19 số 1 )
= ( 1/19 + 1 ) + ( 2/18 + 1) + ( 3/17 +1 ) + ...+ ( 18/2 + 1 ) + 1
= 20/19 + 20/18 + 20/17 + ... + 20/2 + 20/20
= 20 x ( 1/2 + 1/3 + ... + 1/19 + 1/20 )
Vậy phân số trên= 20
23 tháng 5 2016
Mẫu số = 1/19 + 2/18 + 3/17 + ... + 18/2 + 19/1
= ( 1/19 + 2/18 + 3/17 + ... + 18/2 ) + ( 1 + 1 + ... + 1 )
( 18 phân số ) ( 19 số 1 )
= ( 1/19 + 1 ) + ( 2/18 + 1) + ( 3/17 +1 ) + ...+ ( 18/2 + 1 ) + 1
= 20/19 + 20/18 + 20/17 + ... + 20/2 + 20/20
= 20 x ( 1/2 + 1/3 + ... + 1/19 + 1/20 )
Vậy phân số trên= 20
NM
Nguyễn Minh Quang
Giáo viên
25 tháng 1 2021
ta có
tử số \(\frac{1}{19}+\frac{2}{18}+..+\frac{18}{2}+\frac{18}{1}=\frac{1}{19}+1+\frac{2}{18}+1+..+\frac{18}{2}+1\)
\(\frac{20}{19}+\frac{20}{18}+..+\frac{20}{2}=20\left(\frac{1}{19}+\frac{1}{18}+..+\frac{1}{2}\right)\)
Do đó ta có phân số trên bằng 20
14 tháng 9 2016
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
15 tháng 9 2016
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
15 tháng 9 2016
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)