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25S = 1 - 1/52+1/54- 1/56+.......+1/52008- 1/52010
Cộng 2 vế với S ta có :
26S = 1 - 1/52012 < 1 suy ra S< 1/26
\(5^2.S=1-\frac{1}{5^2}+\frac{1}{5^4}-.....+\frac{1}{5^{2008}}-\frac{1}{5^{2010}}\)
\(25S=1-\frac{1}{5^2}+\frac{1}{5^4}-...+\frac{1}{5^{2008}}-\frac{1}{5^{2010}}\)Cộng 2 vế với S ta có
\(26S=1-\frac{1}{5^{2012}}\)\(\Rightarrow26S< 1\Rightarrow S< \frac{1}{26}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Bài 1
\(a,\left|x\right|=-\left|-\frac{5}{7}\right|=>x\in\varnothing\)
\(b,\left|x+4,3\right|-\left|-2,8\right|=0\)
\(=>\left|x+4,3\right|-2,8=0\)
\(=>\left|x+4,3\right|=0+2,8=2,8\)
\(=>x+4,3=\pm2,8\)
\(=>\hept{\begin{cases}x+4,3=2,8\\x+4,3=-2,8\end{cases}=>\hept{\begin{cases}x=-1,5\\x=-7,1\end{cases}}}\)
\(c,\left|x\right|+x=\frac{2}{3}\)
\(=>\hept{\begin{cases}x+x=\frac{2}{3}\\-x+x=\frac{2}{3}\end{cases}}=>\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
1) \(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{13}+\frac{x+4}{12}\)
\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{13}-\frac{x+16}{12}=0\)
\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{13}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x=-16\)
2)3)4) tương tự
Gợi ý : 2) cộng 3 vào cả hai vế
3)4) cộng 2 vào cả hai vế
5) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=-3\)
\(\Leftrightarrow\frac{x+21}{20}+\frac{x+21}{19}+\frac{x+21}{18}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}\right)=0\)
\(\Leftrightarrow x=-21\)
6) sửa VT = 4 rồi tương tự câu 5)
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
\(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(\frac{1}{5}A=\frac{1}{5}\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\right)=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(A-\frac{1}{5}A=\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\left(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\right)\)
\(\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2013}}\)
\(\frac{4}{5}A=\frac{1}{5}-\frac{1}{5^{2013}}=\frac{5^{2012}-1}{5^{2013}}\)
\(A=\frac{5^{2012}-1}{5^{2013}}:\frac{4}{5}=\frac{5^{2012}-1}{5^{2013}}\times\frac{5}{4}=\frac{5^{2012}-1}{4.5^{2012}}=\frac{1}{4}-\frac{1}{4.5^{2012}}< \frac{1}{4}\)