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\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)
\(\left( {\dfrac{1}{7}x - \dfrac{2}{7}} \right)\left( {\dfrac{{ - 1}}{5}x + \dfrac{3}{5}} \right)\left( {\dfrac{1}{3}x + \dfrac{1}{3}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x - \dfrac{2}{7} = 0\\ \dfrac{{ - 1}}{5}x + \dfrac{3}{5} = 0\\ \dfrac{1}{3}x + \dfrac{1}{3} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x = \dfrac{2}{7}\\ - \dfrac{1}{5}x = - \dfrac{3}{5}\\ \dfrac{1}{3}x = - \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 3\\ x = - 1 \end{array} \right. \)
\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)
\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\
\(-2x-\frac{-1}{2}=2x-1\)
\(2x--2x=1-\frac{-1}{2}\)
\(\)\(4x=\frac{3}{2}\)
\(x=\frac{3}{2}:4\)
\(x=\frac{3}{8}\)
P=1/1.2.3.4 +1/2.3.4.5 +1/3.4.5.6 +...+1/97.98.99.100
3P=3/1.2.3.4 +3/2.3.4.5 +3/3.4.5.6 +...+3/97.98.99.100
3P=1/1.2.3-1/2.3.4+1/2.3.4-1/3.4.5+................+1/97.98.99-1/98.99.100
3P = 1/1.2.3 - 1/98.99.100
3P =( 98.99.100-1.2.3)/1.2.3.98.99.100
P=( 98.99.100-1.2.3)/1.2.3.98.99.100.3
P=(98.33.50-1)/98.99.100.3
P= 161699/2910600
=\(\frac{3\left(\frac{1}{1}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{2}{4}+\frac{2}{6}+\frac{2}{8}}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)
=\(\frac{3}{5}+\frac{2\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)=\(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
Ta có \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
Vì \(\frac{9}{10^{12}-1}< \frac{9}{10^{11}+1};1=1\Rightarrow1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\Rightarrow\frac{10^{11}-1}{10^{12}-1}< \frac{10^{10}+1}{10^{11}+1}\)
Suy ra\(A< B\)
\(A=\frac{10^{11}-1}{10^{12}-1}\) => \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}\)
=> \(10A=1-\frac{9}{10^{12}-1}\)=> 10A < 1
\(B=\frac{10^{10}+1}{10^{11}+1}\) => \(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}\)
=> \(10B=1+\frac{9}{10^{11}+1}\)=> 10B > 1
=> 10B > 10A => B > A
ĐS: B > A
1.A= 1.2.3+2.3.4+...+29.30.31+x=15
\(4A=1.2.3.4+2.3.4.\left(5-1\right)+...+29.30.31.\left(32-28\right)+4x=60\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+29.30.31.32-28.29.30.31+4x=60\)
Từ đó suy ra nha bạn
2.\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(=\frac{2}{2\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\\ =1-\frac{2}{\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2009}\Rightarrow x+1=2009\Rightarrow x=2008\)
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