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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
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\dfrac{6}{56}\)
\(A=\dfrac{1}{1}.\dfrac{1}{56}\)
\(A=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+\dfrac{10}{34.44}+\dfrac{10}{44.54}+\dfrac{10}{54.64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{5}{60}-\dfrac{2}{60}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(B=5.\dfrac{3}{60}-\left(\dfrac{4}{24}-\dfrac{4}{64}\right)\)
\(B=5.\dfrac{1}{20}-\left(\dfrac{1}{6}-\dfrac{1}{16}\right)\)
\(B=\dfrac{5}{20}-\left(\dfrac{8}{48}-\dfrac{3}{48}\right)\)
\(B=\dfrac{1}{4}-\dfrac{5}{48}\)
\(B=\dfrac{12}{48}-\dfrac{5}{48}\)
\(B=\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}:\dfrac{7}{48}\)
\(\dfrac{A}{B}=\dfrac{1}{56}.\dfrac{48}{7}\)
\(\dfrac{A}{B}=\dfrac{1}{7}.\dfrac{6}{7}\)
\(\dfrac{A}{B}=\dfrac{6}{49}=\dfrac{48}{392}< \dfrac{49}{392}=\dfrac{1}{8}\)
\(\dfrac{A}{B}< \dfrac{1}{8}\)
Vậy \(\dfrac{A}{B}< \dfrac{1}{8}\)
\(P=\dfrac{1000}{100-x}\)
\(P_{MAX}\Rightarrow P\in Z^+\)
\(\Rightarrow100-x=1\)
\(\Rightarrow x=100-1=99\)
\(\Rightarrow P_{MAX}=\dfrac{1000}{100-99}=1000\)
\(A=\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+.....+\dfrac{1}{50.56}\)
\(A=\dfrac{1}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+.....+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(A=\dfrac{1}{6}.\left(\dfrac{1}{8}-\dfrac{1}{56}\right)=\dfrac{1}{6}.\dfrac{3}{28}=\dfrac{1}{56}\)
\(B=\dfrac{45}{12.21}+\dfrac{45}{21.30}-\dfrac{40}{24.34}-\dfrac{40}{34.44}-\dfrac{40}{44.54}-\dfrac{40}{54.64}\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-5\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(B=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}+\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{54}+\dfrac{1}{54}-\dfrac{1}{64}\right)\)\(B=5\left(\dfrac{1}{12}-\dfrac{1}{64}\right)=5.\dfrac{13}{192}=\dfrac{65}{192}\)
\(\dfrac{A}{B}=\dfrac{1}{\dfrac{56}{\dfrac{65}{192}}}=\dfrac{24}{455}\)
\(\dfrac{1}{8}=\dfrac{3}{24}\)
\(\Rightarrow\dfrac{A}{B}< \dfrac{1}{8}\rightarrowđpcm\)
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) \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}+\dfrac{2}{3}-1\dfrac{15}{17}\)
\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)
\(=1+1-1\dfrac{15}{17}=\dfrac{2}{17}\)
Đề sai:
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1983}+1\right)=0\)
\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên \(2013-x=0\Leftrightarrow x=2013\)
\(\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}+4=0\)
\(\Rightarrow\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1963}+1\right)+\left(\dfrac{40-x}{1968}+1\right)+\left(\dfrac{35-x}{1973}+1\right)+\left(\dfrac{30-x}{1978}+1\right)=-4+1+1+1+1\)
\(\Rightarrow\dfrac{2008-x}{1963}+\dfrac{2008-x}{1968}+\dfrac{2008-x}{1973}+\dfrac{2008-x}{1978}=0\)
Vì \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\) nên 2008 - x = 0
\(\Rightarrow x=2008\)
\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\\ \Leftrightarrow \frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\\ \Leftrightarrow \frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\\ \Leftrightarrow \frac{x}{15}=\frac{y}{20}=\frac{z}{40}\\\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\\ \Rightarrow x=15k;y=20k;z=40k\\ xy=1200\\ \Leftrightarrow 15k.20k=300k^2=1200\\ \Leftrightarrow k^2=4\)
\(\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ k=2\Rightarrow\left\{{}\begin{matrix}x=15k=15.2=30\\y=20k=20.2=40\\z=40k=40.2=80\end{matrix}\right.\\ k=-2\Rightarrow\left\{{}\begin{matrix}x=15k=15.\left(-2\right)=-30\\y=20k=20.\left(-2\right)=-40\\z=40k=40.\left(-2\right)=-80\end{matrix}\right.\)
\(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)
\(\Rightarrow\dfrac{x-9}{15}=\dfrac{y-12}{20}=\dfrac{z-24}{40}=k\)
\(\Rightarrow\left(15k+9\right)\left(20k+12\right)=1200\)
\(\Rightarrow3.4\left(5k+3\right)\left(5k+3\right)=1200\)
\(\Rightarrow\left(5k+3\right)\left(5k+3\right)=1200:3:4\)
\(\Rightarrow\left(5k+3\right)^2=100\)
\(\Rightarrow\left[{}\begin{matrix}5k+3=10\\5k+3=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}k=\dfrac{7}{5}\\k=\dfrac{-13}{5}\end{matrix}\right.\)
+) Với \(k=\dfrac{7}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}.15+9\\y=\dfrac{7}{5}.20+12\\z=\dfrac{7}{5}.40+24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=80\end{matrix}\right.\)
+) Với \(k=\dfrac{-13}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-13}{5}.15+9\\y=\dfrac{-13}{5}.20+12\\z=\dfrac{-13}{5}.40+24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-30\\y=-40\\z=-80\end{matrix}\right.\)
Vậy .................
ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)
\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)
\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)
\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)