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Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Ta có :
\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)
a) Giải
Ta có: \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\)
\(\Rightarrow2S=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{2012}}+\dfrac{2}{2^{2013}}\)
\(2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2S-S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2012}}-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=1-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=\dfrac{2^{2013}-1}{2^{2013}}\)
b) Giải
Từ \(A=\dfrac{2011^{2012}+1}{2011^{2013}+1}\)
\(\Rightarrow2011A=\dfrac{2011^{2013}+20111}{2011^{2013}+1}=\dfrac{2011^{2013}+1+2010}{2011^{2013}+1}=1+\dfrac{2010}{2011^{2013}+1}\)
Từ \(B=\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
\(\Rightarrow2011B=\dfrac{2011^{2014}+2011}{2011^{2014}+1}=\dfrac{2011^{2014}+1+2010}{2011^{2014}+1}=1+\dfrac{2010}{2011^{2014}+1}\)
Vì 20112013 + 1 < 20112014 + 1 và 2010 > 0
\(\Rightarrow\dfrac{2010}{2011^{2013}+1}>\dfrac{2010}{2011^{2014}+1}\)
\(\Rightarrow2011A>2011B\)
\(\Rightarrow A>B\)
Vậy A > B.
a) \(\dfrac{5x-3}{3-2x}=\dfrac{2}{3}\)
\(\Rightarrow3\left(5x-3\right)=2\left(3-2x\right)\)
\(\Rightarrow15x-9=6-4x\)
\(\Rightarrow15x+4x=9+6\)
\(\Rightarrow19x=15\Rightarrow x=\dfrac{15}{19}\)
b) \(\left(\dfrac{4}{5}x+\dfrac{2}{3}\right):\dfrac{3}{4}=2\)
\(\Rightarrow\dfrac{4}{5}x+\dfrac{2}{3}=\dfrac{3}{2}\Rightarrow\dfrac{4}{5}x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{25}{24}\)
c) \(\dfrac{3}{4}x-\dfrac{1}{3}=\dfrac{3}{5}\Rightarrow\dfrac{3}{4}x=\dfrac{14}{15}\)
\(\Rightarrow x=\dfrac{56}{45}\)
d) \(\dfrac{2}{3}-\dfrac{3}{5}:x=\dfrac{1}{4}\Rightarrow\dfrac{3}{5}:x=\dfrac{5}{12}\)
\(\Rightarrow x=\dfrac{36}{25}\)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
Đặt \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)
Ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2008^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2007\cdot2008}\left(1\right)\)
Lại có: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
\(=1-\dfrac{1}{2008}< 1\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\) ta có \(A< B< 1\Rightarrow A< 1\)
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2008^2}\)
A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
A<\(1-\dfrac{1}{2008}\)
A<\(\dfrac{2007}{2008}< 1\)
=> A<1
Vậy A<1
b) \(\dfrac{4}{5}-\dfrac{3}{4}:x=0,3\)
\(\Rightarrow0,8-0,75:x=0,3\)
\(\Rightarrow0,75:x=0,5\)
\(\Rightarrow x=1,5\)
c) \(\dfrac{-3}{2}-\dfrac{1}{4}x=1\dfrac{1}{3}-0,2x\)
\(\Rightarrow\dfrac{-3}{2}-\dfrac{4}{3}=\dfrac{1}{4}x-\dfrac{1}{5}x\)
\(\Rightarrow x=\dfrac{-17}{6}\cdot20\)
\(\Rightarrow x=\dfrac{-170}{3}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{1.3}\)
\(...\)
\(\dfrac{1}{100^2}>\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{100}=\dfrac{99}{100}\\ \dfrac{99}{100}< \dfrac{1}{4}\\ \Rightarrowđpcm\)