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\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
Nếu A= thì
Ta có 2/2^2 + 2/3^3 + 2/4^2 +... + 2/2016^2 + 2/ 2017^2 = 2( 1/ 2^2 + 1/3^2 + 1/ 4^2 +... + 1/2016^2 + 1/2017^2
Mà 2( 1/ 2^2 + 1/3^2 + 1/ 4^2 +... + 1/2016^2 + 1/2017^2 < 2( 1/1.2 + 1/2.3 + 1/ 3.4 + ... + 1/ 2015.2016 + 1/2016 + 2017) = 2( 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +... + 1/2015 - 1/ 2016 + 1/2016 - 1/2017) = 2( 1- 1/2017) = 2( 2016/2017) = 4032 / 2017< 2 => 2( 1/ 2^2 + 1/3^2 + 1/ 4^2 +... + 1/2016^2 + 1/2017^2 < 2 => 2/2^2 + 2/3^3 + 2/4^2 +... + 2/2016^2 + 2/ 2017^2 < 2 => A<2
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{100}{100}-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=99\)
ta có: \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{8}{2^5-3}=\frac{8}{29}=\frac{104}{377}\)
\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{45}{13}=\frac{1305}{377}\)
\(\Rightarrow\frac{104}{377}< \frac{1305}{377}\Rightarrow\frac{2^5.7+2^5}{2^5.2^5-2^5.3}< \frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)
Ta cứ tính ra tử số và mỗi số của từng phân số ra nhé Jerry Gaming:
\(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)= \(\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{2^5.8}{2^5.\left(32-3\right)}=\frac{32.8}{2^5.29}=\frac{32.8}{32.29}=\frac{8}{29}\)
\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)= \(\frac{3^4.5.3^6}{3^8.13}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{9.5}{13}=\frac{45}{13}\)
\(\frac{8}{29}\)và \(\frac{45}{13}\)MSC: 377
Ta có:
\(\frac{8}{29}=\frac{8.13}{29.13}=\frac{104}{377}\)
\(\frac{45}{13}=\frac{45.29}{13.29}=\frac{1305}{377}\)
Vậy quy đồng \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)và \(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)ta được \(\frac{104}{377}\)và \(\frac{1305}{377}\)
Chúc bạn học tốt!
trong câu hỏi tương tự có một bài giốngđè và được giải rồi, bạn xem thử đi
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
A=\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)
A=\(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
A=\(2.\left(1-\frac{1}{13}\right)\)
A=\(2.\frac{12}{13}=\frac{24}{13}\)
A=2(2/1.3+2/3.5+2/5.7+...+2/11.13)
A=2(1/1-1/3+1/3-1/5+1/5-1/7+...+1/11-1/13)
A=2(1/1-1/13)=2.12/13=24/13