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\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\dfrac{6}{56}\)
\(A=\dfrac{1}{1}.\dfrac{1}{56}\)
\(A=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)
\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)
\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)
\(B=\dfrac{1}{4}-\dfrac{5}{48}\)
\(B=\dfrac{12}{48}-\dfrac{5}{48}\)
\(B=\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)
\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)
\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)
\(P=\dfrac{1000}{100-x}\)
\(P_{MAX}\Rightarrow P\in Z^+\)
\(\Rightarrow100-x=1\)
\(\Rightarrow x=100-1=99\)
\(\Rightarrow P_{MAX}=\dfrac{1000}{100-99}=1000\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+.....+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+.....+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)=\dfrac{1}{6}.\dfrac{3}{28}=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-5\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}+\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{1}{12}-\dfrac{1}{64}\right)=5.\dfrac{13}{192}=\dfrac{65}{192}\)
\(\dfrac{A}{B}=\dfrac{1}{\dfrac{56}{\dfrac{65}{192}}}=\dfrac{24}{455}\)
\(\dfrac{1}{8}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{A}{B}< \dfrac{1}{8}\rightarrowđpcm\)
B=5(1/12−1/21+1/21−1/30)−5(1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64)
B=5(1/12−1/21+1/21−1/30+1/24−1/34+1/34−1/44+1/44−1/54+1/54−1/64 )
B=5(1/12−1/64)=5.13/192=65/192
chứng minh rằng:\(\frac{1}{5^3}+\frac{1}{6^3}+....+\frac{1}{2016^3}+\frac{1}{2017^3}< \frac{1}{40}\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)
\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\frac{3}{28}\)
\(A=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)
\(B=5.\frac{1}{20}-4.\frac{5}{192}\)
\(B=\frac{1}{4}-\frac{5}{48}\)
\(B=\frac{7}{48}\)
Ta có \(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}=\frac{1}{56}\times\frac{48}{7}=\frac{6}{49}\)
Lấy \(\frac{6}{49}-\frac{1}{8}=-\frac{1}{392}< 0\)
\(\Rightarrow\frac{6}{49}< \frac{1}{8}\) hay \(\frac{A}{B}< \frac{1}{8}\)
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+....+\frac{1}{50.56}\)
\(=\frac{1}{6}.(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+....+\frac{6}{50.56})\)
\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+....+\frac{1}{50}-\frac{1}{56})\)
\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{56})\)
\(=\frac{1}{6}.(\frac{7}{56}-\frac{1}{56})\)
\(=\frac{1}{6}.\frac{6}{56}\)
\(=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(=5(\frac{9}{12.21}+\frac{9}{21.30})-4(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64})\)
\(=5(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64})\)
\(=5(\frac{1}{12}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{64})\)
\(=5(\frac{5}{60}-\frac{2}{60})-(\frac{4}{24}-\frac{4}{64})\)
\(=5.\frac{1}{20}-(\frac{1}{6}-\frac{1}{16})\)
\(=\frac{1}{4}-(\frac{8}{48}-\frac{3}{48})\)
\(=\frac{1}{4}-\frac{5}{48}\)
\(=\frac{12}{48}-\frac{5}{48}=\frac{7}{48}\)
\(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}\)
\(=\frac{1}{56}.\frac{48}{7}\)
\(=\frac{6}{49}=\frac{48}{392}\)bé hơn \(\frac{49}{392}=\frac{1}{8}\)
Vậy \(\frac{A}{B}\)bé hơn \(\frac{1}{8}\)
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