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\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.......+\frac{15}{74.77}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+.......+\frac{3}{74.77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+.....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=5\left(\frac{7}{77}-\frac{1}{77}\right)\)
\(=5.\frac{6}{77}\)
\(=\frac{30}{77}\)
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+...+\frac{15}{72.75}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{72.75}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+...+\frac{1}{72}-\frac{1}{75}\right)\)\(=5\left(\frac{1}{11}-\frac{1}{75}\right)\)
\(=\frac{64}{165}\)
\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{68.71}\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{68}-\dfrac{1}{71}\right)\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{71}\right)\)
\(=5.\dfrac{60}{781}\)
\(=\dfrac{300}{781}\)
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
\(=\left(\dfrac{8}{9}+\dfrac{1}{9}\right)+\left(\dfrac{-15}{23}+\dfrac{15}{23}\right)+\dfrac{1}{2}\)
\(=1+0+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
=\((\dfrac{8}{9}+\dfrac{1}{9})\) +\((\dfrac{15}{23}+\dfrac{-15}{23})\) +\(\dfrac{1}{2}\)
= 1+ 0+\(\dfrac{1}{2}\)
= \(\dfrac{3}{2}\)
a, \(\left(\dfrac{-5}{11}\right).\dfrac{7}{15}.\left(\dfrac{11}{-5}\right).\left(-30\right)=\dfrac{-5}{11}.\dfrac{7}{15}.\dfrac{-11}{5}.\dfrac{-30}{1}\)= ( - 14 )
b, \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{1.4.3}{3.3.5}=\dfrac{4}{15}\)
c, \(\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{15}{-7}.\left(-16\right)=\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{-15}{7}.\dfrac{-16}{1}\)
\(\dfrac{-1.5.-1.-2}{1.1.1.1}=\left(-10\right)\)
d,\(\left(\dfrac{-1}{2}\right).3\dfrac{1}{5}+\left(\dfrac{-1}{2}\right).-2\dfrac{1}{5}=\left(\dfrac{-1}{2}\right).\left[\dfrac{16}{5}+\left(\dfrac{-11}{5}\right)\right]\)
= \(\left(\dfrac{-1}{2}\right).1=\dfrac{-1}{2}\)
a) (-5/11.11/-5).7/15
=1.7/15=7/15
b)(11/12:33/16).3/5
=(11/12.16/33).3/5
=4/9.3/5=4/15
c)(-7/15.15/-7).5/8
=1.5/8=5/8
d)(-1/2).(16/5.-11/5)
=-1/2.1=-1/2
xg r đó
\(1,\dfrac{2x+4}{7}=\dfrac{4x-2}{15}=\dfrac{2.\left(2x+4\right)}{2.7}=\dfrac{4x+8}{14}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}==\dfrac{4x+8}{14}=\dfrac{\left(4x+8\right)-\left(4x-2\right)}{14-15}=\dfrac{10}{-1}=-10\)
\(\Rightarrow\dfrac{2x+4}{7}=-10\)
\(\Rightarrow2x+4=-10.7=-70\)
\(\Rightarrow2x=-70+4=-66\)
\(\Rightarrow x=-66:2=-33\)
Vậy \(x=-33\)
\(2,\dfrac{2x+3}{5}=\dfrac{7x-3}{15}=\dfrac{7.\left(2x+3\right)}{7.5}=\dfrac{2.\left(7x-3\right)}{2.15}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+3}{5}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}=\dfrac{\left(14x+21\right)-\left(14x-6\right)}{35-30}=\dfrac{29}{5}\)
\(\Rightarrow\dfrac{2x+3}{5}=\dfrac{29}{5}\)
\(\Rightarrow2x+3=29\)
\(\Rightarrow2x=29-3=26\)
\(\Rightarrow x=26:2=13\)
\(3,\dfrac{11x-2}{7x+5}=\dfrac{11}{8}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{7x+5}{8}=\dfrac{7.\left(11x-2\right)}{7.11}=\dfrac{11.\left(7x+5\right)}{8.11}=\dfrac{77x-14}{77}=\dfrac{77x+55}{88}=\dfrac{\left(77x+55\right)-\left(77x-14\right)}{88-77}=\dfrac{69}{11}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{69}{11}\)
\(\Rightarrow11x-2=69\)
\(\Rightarrow11x=69+2=71\)
\(\Rightarrow x=\dfrac{71}{11}\)
\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)
\(=\dfrac{18+12+\left(-5\right)}{24}\)
\(=\dfrac{25}{24}\)
\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)
\(=\dfrac{5}{7}.0=0\)
\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)
\(=\dfrac{1391}{714}\)
a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)
b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)
c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)
d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
theo bài ra ta có:
\(E=\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{74.77}\\ \Rightarrow\dfrac{1}{5}E=\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{74.77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{74}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{7}{77}-\dfrac{1}{77}=\dfrac{6}{77}\\ \Rightarrow E=\dfrac{6}{77}.5\\ E=\dfrac{30}{77}\)
5 .\((\)\(\dfrac{3}{11.14}+\dfrac{3}{14.17}+...+\dfrac{3}{74.77}\))
= 5. (\(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{74}-\dfrac{1}{77}\))
= 5.(\(\dfrac{1}{11}-\dfrac{1}{77}\))
= 5. \(\dfrac{6}{77}\)
= \(\dfrac{30}{77}\)