1/1x2 + 1/2x3 + 1/3x4 + ... + 1/X x (x+1) = 1999/2000
1/1x2 + 1/2x3 + 1/3x4 + ... 1/99 x 100
1/1.2 +1/2.3 +1/3.4 +....+1/99.100
=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100
=1-1/100
=99/100
tham khảo
1/1.2 +1/2.3 +1/3.4 +....+1/99.100
=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100
=1-1/100
=99/100
Tìm x biết 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/X(X+1) = 99/100
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
1/(1x2)+1/(2x3)+1/(3x4)...+1/xx(x+1)
Ta có : A = \(\frac{1}{1\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{3\text{x}4}+...+\frac{1}{X\text{x}\left(X+1\right)}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
A = \(\frac{1}{1}-\frac{1}{x+1}\)
A = \(\frac{x}{x+1}\)
Ủng hộ mik nhá !!!!
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-?\)
\(\Rightarrow x+1=?\Leftrightarrow x=?\)
x/1x2 +x/2x3 + x/3x4+ .....+x/2017x2018=-1
x= -2018/2017
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)
\(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{2017\cdot2018}=-1\)
=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2017}-\frac{x}{2018}=-1\)
=> \(\frac{x}{1}-\frac{x}{2018}=-1\)
=> \(\frac{2018x-x}{2018}=-1\)
=> \(\frac{2017x}{2018}=-1\)
=> 2017x = -2018
=> x = -2018/2017
1/1x2+1/2x3+1/3x4+1/4x5+...+1/X nhân ( X + 1 ) = 2017/2018 làm ơn đó
1x2+2x3+3x4+...+x(x+1)
giúp mih với
Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này?
tìm x biết
a, (1/1x2+1/2x3+1/5x4+...+1/99x100) X=1/1x2+2x3+3x4+...+98x99
b, X/1x3+X/3x5+X/5x7+...+X/2013x2015=4/2015
c, X+1/2015+X+2/2016=X+3/2017+X+4/2018
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
tính tổng: 1/1x2 + 1/2x3 + 1/ 3x4 +...+ 1/98 x 99 + 1/99 x 100
1/1*2 + 1/2*3 + 1/3*4 + .... + 1/99 * 100
= 1- 1/100
= 99/100
=1-1/2+1/2-...-1/99+1/99-1/100=1-1/100=99/100
Tìm X:
1:1x2+1:2x3+1:3x4+....+1: (X-1)x X=15:16
Chú ý : dấu chia là phân số
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{\left(x-1\right)\times x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{x}=\dfrac{15}{16}\)
\(\dfrac{1}{x}=1-\dfrac{15}{16}=\dfrac{16}{16}-\dfrac{15}{16}\)
\(\dfrac{1}{x}=\dfrac{1}{16}\)
\(\Rightarrow x=16\)
Tính tổng sau : 1/1x2 + 1/2x3 + 1/3x4 ..... + 1/2009 x 2010
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)