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\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{68.71}\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{68}-\dfrac{1}{71}\right)\)
\(=5\left(\dfrac{1}{11}-\dfrac{1}{71}\right)\)
\(=5.\dfrac{60}{781}\)
\(=\dfrac{300}{781}\)
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. \(\dfrac{6}{77}\)
= \(\dfrac{30}{77}\)
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+.......+\frac{15}{74.77}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+.......+\frac{3}{74.77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+.....+\frac{1}{74}-\frac{1}{77}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{77}\right)\)
\(=5\left(\frac{7}{77}-\frac{1}{77}\right)\)
\(=5.\frac{6}{77}\)
\(=\frac{30}{77}\)
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
\(=\left(\dfrac{8}{9}+\dfrac{1}{9}\right)+\left(\dfrac{-15}{23}+\dfrac{15}{23}\right)+\dfrac{1}{2}\)
\(=1+0+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
\(\dfrac{8}{9}+\dfrac{15}{23}+\dfrac{1}{9}+\dfrac{-15}{23}+\dfrac{1}{2}\)
=\((\dfrac{8}{9}+\dfrac{1}{9})\) +\((\dfrac{15}{23}+\dfrac{-15}{23})\) +\(\dfrac{1}{2}\)
= 1+ 0+\(\dfrac{1}{2}\)
= \(\dfrac{3}{2}\)
a. = 1/20 + 5 - 1/2
= 101/20 - 1/2
= 91/20
b. = ( 6/15 - 3/5) - ( 7/8 + 2/16) + 3
= -1/5 - 1 + 3
= 9/5
c. = 15/7 . ( 3/5 - 8/5)
= 15/7 . ( -1)
= - 15/7
e. = -14/9 - 3/9
= -17/9
f. = 19/21 . ( 15/17 + 2/17) + 13/21
= 19/21 . 1 + 13/21
= 32/21
g. = 43/12 : 2 + 5/24
= 43/24 + 5/24
= 2
\(\frac{3}{15}\cdot G=\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{68\cdot71}\)
\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)
\(\frac{3}{15}\cdot G=\frac{1}{11}-\frac{1}{71}\)
\(G=\frac{60}{781}\cdot\frac{15}{3}\)
\(G=\frac{300}{781}\)
ta có :\(\frac{3}{15}G=\left(\frac{15}{11.14}+\frac{15}{14.17}+...+\frac{15}{68.71}\right)\)
\(\frac{3}{15}G=\frac{3}{11.14}+\frac{3}{14.17}+...+\frac{3}{68.71}\)
\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{68}-\frac{1}{71}\)
\(\frac{3}{15}G=\frac{1}{11}-\frac{1}{71}=\frac{71}{781}-\frac{11}{781}=\frac{60}{781}\)
\(=>G=\frac{60}{781}:\frac{3}{15}=\frac{900}{2343}\)
vậy G =900/2343
a: \(A=\dfrac{-13}{21}=\dfrac{-26}{42}\)
\(B=\dfrac{-9}{14}=\dfrac{-27}{42}\)
mà -26>-27
nên A>B
b: \(A=\dfrac{99}{101}=1-\dfrac{2}{101}\)
\(B=\dfrac{2011}{2013}=1-\dfrac{2}{2013}\)
mà 2/101>2/2013
nên A<B
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
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
\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+...+\frac{15}{72.75}\)
\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{72.75}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+...+\frac{1}{72}-\frac{1}{75}\right)\)\(=5\left(\frac{1}{11}-\frac{1}{75}\right)\)
\(=\frac{64}{165}\)
pài này gần giống pài troq v15