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\(A=_{100}+5+5.9+5.9^2+5.9^3+...+5.9^{99}\)
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ND
Nguyễn Đăng Nhân
CTVHS
25 tháng 8 2023
\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)
\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)
\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)
\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)
ND
Nguyễn Đức Trí
VIP
25 tháng 8 2023
\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)
\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)
\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}\)
PT
0
HC
13 tháng 3 2017
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AI TK MÌNH MÌNH TK LẠI 3 LẦN.
8 tháng 9 2016
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(A=\frac{2^{10}.3^8.\left(-2\right)}{2^{10}.3^8.6}\)
\(A=-\frac{2^{11}.3^8}{2^{10}.3^8.2.3}=-\frac{2^{11}.3^8}{2^{11}.3^9}=-\frac{1}{3}\)
25 tháng 2 2017
2.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{2}.\left(\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow\)2x + 3 = 93
\(\Rightarrow\)2x = 93 - 3
\(\Rightarrow\)2x = 90
\(\Rightarrow\)x = 90 : 2 = 45
S
25 tháng 2 2017
\(H=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{33.37}\)
= \(\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{33}-\frac{1}{37}\right)\)
= \(\frac{3}{4}\left(1-\frac{1}{37}\right)\)
= \(\frac{3}{4}.\frac{36}{37}=\frac{27}{37}\)
20 tháng 12 2022
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
