Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\dfrac{1000}{100-x}\)
\(P_{MAX}\Rightarrow P\in Z^+\)
\(\Rightarrow100-x=1\)
\(\Rightarrow x=100-1=99\)
\(\Rightarrow P_{MAX}=\dfrac{1000}{100-99}=1000\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+.....+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+.....+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)=\dfrac{1}{6}.\dfrac{3}{28}=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-5\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}+\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{1}{12}-\dfrac{1}{64}\right)=5.\dfrac{13}{192}=\dfrac{65}{192}\)
\(\dfrac{A}{B}=\dfrac{1}{\dfrac{56}{\dfrac{65}{192}}}=\dfrac{24}{455}\)
\(\dfrac{1}{8}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{A}{B}< \dfrac{1}{8}\rightarrowđpcm\)
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)
\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\frac{3}{28}\)
\(A=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)
\(B=5.\frac{1}{20}-4.\frac{5}{192}\)
\(B=\frac{1}{4}-\frac{5}{48}\)
\(B=\frac{7}{48}\)
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 \(\frac{1}{8}\)
Chúc bạn học tốt
ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)
\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)
\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)
\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)
a: \(\dfrac{8}{18}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{4+15}{9}=\dfrac{19}{9}\)
b: \(\dfrac{8}{24}+\dfrac{4}{48}=\dfrac{1}{3}+\dfrac{1}{12}=\dfrac{4}{12}+\dfrac{1}{12}=\dfrac{4+1}{12}=\dfrac{5}{12}\)
c: \(\dfrac{20}{15}-\dfrac{4}{45}=\dfrac{4}{3}-\dfrac{4}{45}=\dfrac{60}{45}-\dfrac{4}{45}=\dfrac{60-4}{45}=\dfrac{56}{45}\)
d: \(\dfrac{40}{32}-\dfrac{1}{2}=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{5-2}{4}=\dfrac{3}{4}\)
\(a,\dfrac{-5}{13}+\dfrac{8}{13}=\dfrac{3}{13}\\ b,\dfrac{5}{31}+\dfrac{-22}{31}=\dfrac{-17}{31}\\ c,\dfrac{-13}{43}+\dfrac{-40}{43}=\dfrac{-53}{43}\\ d,\dfrac{-3}{29}-\dfrac{16}{58}=\dfrac{-11}{29}\\ e,\dfrac{8}{40}-\dfrac{-36}{45}=1\\ f,\dfrac{-8}{18}-\dfrac{-15}{27}=\dfrac{1}{9}\\ g,\left(-2\right)+\dfrac{-5}{8}=\dfrac{-21}{8}\)
\(-45\%x+2\dfrac{3}{8}=-1\dfrac{31}{40}\)
\(\Rightarrow-\dfrac{9}{20}x+\dfrac{19}{8}=-\dfrac{71}{40}\)
\(\Rightarrow-\dfrac{9}{20}x=-\dfrac{71}{40}-\dfrac{19}{8}\)
\(\Rightarrow-\dfrac{9}{20}x=-\dfrac{83}{20}\)
\(\Rightarrow x=-\dfrac{83}{20}:\left(-\dfrac{9}{20}\right)\)
\(\Rightarrow x=\dfrac{83}{9}\)
Vậy \(x=\dfrac{83}{9}\)
\(-45\%x+2\dfrac{3}{8}=-1\dfrac{31}{40}\)
\(\Leftrightarrow\dfrac{9}{20}x+\dfrac{19}{8}=-\dfrac{71}{40}\)
\(\Leftrightarrow\dfrac{9}{20}x=-\dfrac{71}{40}-\dfrac{19}{8}\)
\(\Leftrightarrow\dfrac{9}{20}x=-\dfrac{83}{20}\)
\(\Leftrightarrow x=-\dfrac{83}{20}:\dfrac{9}{20}\)
\(\Leftrightarrow x=-\dfrac{83}{20}.\dfrac{20}{9}\)
\(\Leftrightarrow x=-\dfrac{83}{9}\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\dfrac{6}{56}\)
\(A=\dfrac{1}{1}.\dfrac{1}{56}\)
\(A=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)
\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)
\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)
\(B=\dfrac{1}{4}-\dfrac{5}{48}\)
\(B=\dfrac{12}{48}-\dfrac{5}{48}\)
\(B=\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)
\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)
\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)