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a) Đặt \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=k\)
\(\Rightarrow k=\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\frac{x+2y+z}{9a}\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{k}{9}\)
Tương tự :\(\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}=\frac{k}{9}\)
Vậy ..........
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
\(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(\frac{1}{5}A=\frac{1}{5}\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\right)=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(A-\frac{1}{5}A=\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\left(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\right)\)
\(\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2013}}\)
\(\frac{4}{5}A=\frac{1}{5}-\frac{1}{5^{2013}}=\frac{5^{2012}-1}{5^{2013}}\)
\(A=\frac{5^{2012}-1}{5^{2013}}:\frac{4}{5}=\frac{5^{2012}-1}{5^{2013}}\times\frac{5}{4}=\frac{5^{2012}-1}{4.5^{2012}}=\frac{1}{4}-\frac{1}{4.5^{2012}}< \frac{1}{4}\)
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
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}+\frac{6}{7}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{-3}{4}+\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{-5}{6}+\frac{5}{6}\right)\)
\(=\frac{13}{7}+0+0+0+0\)
\(=\frac{13}{7}\)
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}.\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)-\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)-\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{6}{7}\)
\(=1+0-0+0+\frac{6}{7}\)
\(=1+\frac{6}{7}=1\frac{6}{7}\)
d) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=-3+\frac{3}{2}+1-5+\frac{6}{5}-\frac{3}{2}\)
\(=-7+\frac{6}{5}=-\frac{29}{5}\)
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
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
\(26S=1-\frac{1}{5^{2012}}\)\(\Rightarrow26S< 1\Rightarrow S< \frac{1}{26}\)