Rút gọn M=\(\frac{a^{30}+a^{20}+a^{10}+1}{a^{2042}+a^{2032}+a^{2022}+a^{2012}+a^{30}+a^{20}+a^{10}+1}\)
K
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2 tháng 2 2022
\(=\dfrac{a^{20}\left(a^{10}+1\right)+\left(a^{10}+1\right)}{\left(a^{10}+1\right)\left(a^{2032}+a^{2012}+a^{20}+1\right)}\)
\(=\dfrac{a^{20}+1}{\left(a^{20}+1\right)\left(a^{2012}+1\right)}=\dfrac{1}{a^{2012}+1}\)
DN
2
TB
1
26 tháng 1 2022
b: 12/20+6/18=3/5+1/3=9/15+5/15=14/15
c: 8/12+20/30=2/3+2/3=4/3
d: 10/12+12/36=5/6+2/6=7/6
S
12 tháng 1 2020
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}\right)=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)\)
\(=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...-\frac{1}{10}\right)=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{10}\right)=\frac{1}{5}-\frac{1}{4}=\frac{-1}{20}\)
L
12 tháng 1 2020
\(A=\frac{1}{10}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\)
\(A=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=\frac{1}{10}-\left(\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{1}{10}-\left[\left(\frac{1}{4}-\frac{1}{10}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{9}-\frac{1}{9}\right)\right]\)
\(A=\frac{1}{10}-\frac{1}{4}+\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{4}\)
\(A=-\frac{1}{20}\)
NT
1
G
19 tháng 3 2018
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)
\(M=\frac{a^{30}+a^{20}+a^{10}+1}{a^{2012}\left(a^{30}+a^{20}+a^{10}+1\right)+\left(a^{30}+a^{20}+a^{10}+1\right)}\)
\(M=\frac{1}{a^{2012}+1}\)
\(\frac{a^{30}+a^{20}+a^{10}+1}{a^{2042}+a^{2032}+a^{2022}+a^{2012}+a^{30}+a^{20}+a^{10}+1}=\frac{a^{30}+a^{20}+a^{10}+1}{a^{2042}+a^{2032}+a^{2022}+a^{2012}}+1=\frac{1}{a^{2012}}+1\)
=\(\frac{a^{2012}+1}{a^{2012}}\)