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\(P=\dfrac{1000}{100-x}\)
\(P_{MAX}\Rightarrow P\in Z^+\)
\(\Rightarrow100-x=1\)
\(\Rightarrow x=100-1=99\)
\(\Rightarrow P_{MAX}=\dfrac{1000}{100-99}=1000\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+.....+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+.....+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)=\dfrac{1}{6}.\dfrac{3}{28}=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-5\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}+\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{1}{12}-\dfrac{1}{64}\right)=5.\dfrac{13}{192}=\dfrac{65}{192}\)
\(\dfrac{A}{B}=\dfrac{1}{\dfrac{56}{\dfrac{65}{192}}}=\dfrac{24}{455}\)
\(\dfrac{1}{8}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{A}{B}< \dfrac{1}{8}\rightarrowđpcm\)
\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)
\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)
\(A=\frac{1}{6}.\frac{3}{28}\)
\(A=\frac{1}{56}\)
\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)
\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)
\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)
\(B=5.\frac{1}{20}-4.\frac{5}{192}\)
\(B=\frac{1}{4}-\frac{5}{48}\)
\(B=\frac{7}{48}\)
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) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
a) \(\dfrac{x}{-9}=\dfrac{-40}{45}\)
\(\Leftrightarrow x.45=\left(-9\right).\left(-40\right)\)
\(\Leftrightarrow x.45=360\)
\(\Leftrightarrow x=\dfrac{360}{45}=8\)
Vậy x=8
b: \(\Leftrightarrow\left|2x+\dfrac{1}{3}\right|+\dfrac{4}{9}=-5+\dfrac{4}{9}\)
=>|2x+1/3|=-5(vô lý)
a: \(\Leftrightarrow x=\dfrac{-40\cdot\left(-9\right)}{45}=8\)
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\dfrac{6}{56}\)
\(A=\dfrac{1}{1}.\dfrac{1}{56}\)
\(A=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)
\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)
\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)
\(B=\dfrac{1}{4}-\dfrac{5}{48}\)
\(B=\dfrac{12}{48}-\dfrac{5}{48}\)
\(B=\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)
\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)
\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)