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\(\frac{5}{2}+\frac{4}{11}+\frac{1}{11}+\frac{1}{30}+\frac{13}{60}=\frac{5}{2}+\frac{1}{30}+\frac{13}{60}+\frac{4}{11}+\frac{1}{11}\)
\(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}+\frac{5}{11}=\frac{33}{12}+\frac{5}{11}\)
\(\frac{363}{132}+\frac{60}{132}=\frac{423}{132}=\frac{36}{11}\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
A = \(\dfrac{3}{3\times7}\)+ \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\)+...+\(\dfrac{3}{107\times111}\)
A = \(\dfrac{3}{4}\) \(\times\)( \(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+ \(\dfrac{4}{11\times15}\)+...+\(\dfrac{4}{107\times111}\))
A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{107}\)- \(\dfrac{1}{111}\))
A = \(\dfrac{3}{4}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{111}\))
A = \(\dfrac{9}{37}\) > \(\dfrac{9}{45}\) = \(\dfrac{1}{5}\)
Vậy \(\dfrac{3}{3\times7}\) + \(\dfrac{3}{7\times11}\)+ \(\dfrac{3}{11\times15}\) + ...+ \(\dfrac{3}{107\times111}\) > \(\dfrac{1}{5}\) ( đpcm)
Bạn ơi thế này thì đúng hơn chứ:
\(\dfrac{3}{3.7}+\dfrac{3}{7.11}+\dfrac{3}{11.15}+...+\dfrac{3}{107.111}>\dfrac{1}{5}\)
\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=2\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\dfrac{1}{7}A=\dfrac{1}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)
\(=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(=\dfrac{7-2}{2.7}+\dfrac{11-7}{7.11}+\dfrac{14-11}{11.14}+\dfrac{15-14}{14.15}+\dfrac{28-15}{15.28}\)
\(=\dfrac{7}{2.7}-\dfrac{2}{2.7}+\dfrac{11}{7.11}-\dfrac{7}{7.11}+\dfrac{14}{11.14}-\dfrac{11}{11.14}+\dfrac{15}{14.15}-\dfrac{14}{14.15}+\dfrac{28}{15.28}-\dfrac{15}{15.28}\)
\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)
\(=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{14}{28}-\dfrac{1}{28}=\dfrac{13}{28}\)
\(A=\dfrac{13}{28}\div\dfrac{1}{7}=\dfrac{13}{4}\)
Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(\Rightarrow\dfrac{1}{7}.A=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(\Rightarrow\dfrac{1}{7}.A=\left(\dfrac{1}{2}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{14}\right)+\left(\dfrac{1}{14}-\dfrac{1}{15}\right)+\left(\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(\Rightarrow\dfrac{1}{7}.A=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)
\(\Leftrightarrow A=\dfrac{13}{4}\)
Vậy...................