Tìm x:
(\(\frac{3}{1.2}+\frac{3}{3.4}+\frac{3}{5.6}+...+\frac{3}{299.300}\)) X ( \(x+\frac{2}{3}\)) = \(\frac{2}{151}+\frac{2}{152}+\frac{2}{153}+...+\frac{2}{300}\)
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Tìm x (\(\frac{3}{1.2}+\frac{3}{3.4}+\frac{3}{5.6}+...+\frac{3}{299.300}\)) x (\(x+\frac{2}{3}\)) = \(\frac{2}{151}+\frac{2}{152}+...+\frac{2}{300}\)
Tính giúp tớ với \(\frac{3}{1.2}+\frac{3}{3.4}+\frac{3}{5.6}+...+\frac{3}{299.300}\)
=3.(1/1.2+1/2.3+...+1/299.300)
=3.(1-1/2+1/2-1/3+...+1/299-1/300)
=3.(1-1/300)
=3.299/300
=299/100
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1.Tìm số hữu tỉ x:a)frac{x+1}{10}+frac{x+1}{11}+frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}b)frac{x+4}{2000}+frac{x+3}{2001}frac{x+2}{2002}+frac{x+1}{2003}2.CMR:a)frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{49.50}frac{1}{26}+frac{1}{27}+frac{1}{28}+...+frac{1}{50}b)Cho Afrac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{99.100} Chứng minh rằng : frac{7}{12} A frac{5}{6}
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1.Tìm số hữu tỉ x:
a)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b)\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
2.CMR:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
b)Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
Chứng minh rằng : \(\frac{7}{12}< A< \frac{5}{6}\)
Thank you ... Thank you ... Thank .... Thank SOOOOOO MUUUCHHH !!!!!!!!!
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Này! Câu thứ 2, đáp án : Vậy x = -2004
Mà cậu ghi là Vậy x=2001 đó!!
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Tính nhanh:
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.\frac{5^2}{5.6}\)
\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}.\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot\frac{5^2}{5\cdot6}\)
\(=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6}\)
\(=\frac{1}{6}\)
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a)Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{99.100} 1b)Bfrac{1}{3}+left(frac{1}{3}right)^2+left(frac{1}{3}right)^3+...+left(frac{1}{3}right)^{100} frac{1}{2}c)frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{49.50}frac{1}{26}+frac{1}{27}+...+frac{1}{50}d)Afrac{1}{1.2}+frac{1}{3.4}+...+frac{1}{99.100}.CMRfrac{7}{12} A frac{5}{6}AI ĐÚNG MINK left(TICKright)CHO (làm đc trên 2 câu)
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a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
b)B=\(\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{100}< \frac{1}{2}\)
c)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
d)A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}.CMR\frac{7}{12}< A< \frac{5}{6}\)
AI ĐÚNG MINK \(\left(TICK\right)\)CHO (làm đc trên 2 câu)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
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b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
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c)\(C=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
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\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+....+\frac{3}{15.16}\)
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)
=> \(\frac{1}{x}=3\)
=> \(x=\frac{1}{3}\)
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Tính: \(\frac{2}{151}+\frac{2}{152}+\frac{2}{153}+....+\frac{2}{300}\)
So sánh \(A=\frac{2^{2019}}{2^{2019}+3^{2019}}+\frac{3^{2019}}{3^{2019}+5^{2019}}+\frac{5^{2019}}{5^{2019}+2^{2019}}\)với \(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
Tham khảo
https://hoc24.vn/hoi-dap/question/814814.html
B=11.2+13.4+15.6+....+12019.2020
⇒2B=21.2+23.4+25.6+....+22019.2020
<1+12.3+13.4+14.5+15.6+....+12018.2019+12019.2020
2B<1+3−22.3+4−33.4+5−44.5+....+2019−20182018.2019+2020−20192019.2020
2B<1+12−13+13−14+...+12019−12020
2B<1+12−12020<1+12
B<34
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Đặt 22018=a;32019=b;52020=c(a,b,c>0)
A=aa+b+bb+c+cc+a>aa+b+c+ba+b+c+ca+b+c=1
⇒A>1>34>B
1. Chứng Minh Rằng \(\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+.....+\frac{100}{3^{100}}<\frac{3}{4}\)
2. Chứng Minh Rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2012}\)
2.
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\)
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