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a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=\dfrac{1}{1}-\dfrac{1}{20}\)
\(A=\dfrac{20}{20}-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
A = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{18.19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{18.19}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)
A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)
A = \(\dfrac{189}{380}\)
(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha 
)
hôm qua cô giảng cho mình bài này không cần tính đâu
Gọi tổng là A
A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{17.18.19}\)
2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{17.18.19}\)
2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{17.18}-\dfrac{1}{18.19}\)
2A=\(\dfrac{1}{2}-\dfrac{1}{18.19}\)
A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{18.19}\right)\)
A=\(\dfrac{1}{2}.\dfrac{18.19-2}{2.18.19}\) < \(\dfrac{1}{4}\)
A=\(\dfrac{18.19-2}{2.2.18.19}\) < \(\dfrac{18.19}{2.2.18.19}\)
\(\Rightarrow\) A<\(\dfrac{1}{4}\)
\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)
2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))
2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)
2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)
A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)
Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)
Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )
Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)
A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)
A=\(\dfrac{1}{1}-\dfrac{1}{39}\)
A=\(\dfrac{38}{39}\)
còn lại tự làm do mình có việc chút
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
A=\(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\)
5A=\(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+...+\dfrac{5}{5^{2014}}\)
5A=\(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\)
5A-A=\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}\right)\)4A=\(1-\dfrac{1}{5^{2014}}\)
4A=\(\dfrac{5^{2014}-1}{5^{2014}}\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}:4\)
A=\(\dfrac{5^{2014}-1}{5^{2014}}.\dfrac{1}{4}\)
\(\Rightarrow\)A<\(\dfrac{1}{4}\)
Ta có:
A = \(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) 5A = 5\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\)
\(\Rightarrow\) 5A = \(\dfrac{5}{5}+\dfrac{5}{5^2}+\dfrac{5}{5^3}+....+\dfrac{5}{5^{2014}}\)
\(\Rightarrow\) 5A = \(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\)
\(\Rightarrow\)\(\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+....+\dfrac{1}{5^{2013}}\right)\)-\(\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2014}}\right)\) = 5A - A
\(\Rightarrow\)4A= 1 - \(\dfrac{1}{5^{2014}}\)
\(\Rightarrow\) A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Vậy A =\(\dfrac{5^{2014}-1}{5^{2014}}\) : 4
Câu hỏi của Dung Van - Toán lớp 6 | Học trực tuyến tìm kĩ trước khi hỏi nhé.
\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt
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