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\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
\(=\frac{47}{53}.\frac{17}{3}-\frac{47}{53}.\frac{53}{47}+\frac{17}{3}.\frac{6}{17}-\frac{17}{3}.\frac{47}{53}\)
\(=\left(\frac{47}{53}.\frac{17}{3}-\frac{17}{3}.\frac{47}{53}\right)+\left(\frac{47}{53}.\frac{53}{47}-\frac{17}{3}.\frac{6}{17}\right)\)
\(=0+\left(-1\right)\)
\(=-1\).
\(\frac{17}{21}\cdot\frac{48}{53}+\frac{17}{21}\cdot\frac{4}{53}+\frac{52}{53}\cdot\frac{4}{21}\)
\(=\frac{17}{21}\left(\frac{48}{53}+\frac{4}{53}\right)+\frac{52}{53}\cdot\frac{4}{21}\)
\(=\frac{17}{21}\cdot\frac{52}{53}+\frac{52}{53}\cdot\frac{4}{21}\)
\(=\frac{52}{53}\left(\frac{17}{21}+\frac{4}{21}\right)\)
\(=\frac{52}{53}\cdot1\)
\(\frac{17}{21}.\frac{48}{53}+\frac{17}{21}.\frac{4}{53}+\frac{52}{53}.\frac{4}{21}\)
= \(\frac{17}{21}.\left(\frac{48}{53}+\frac{4}{53}\right)+\frac{52}{53}.\frac{4}{21}\)
= \(\frac{17}{21}.\frac{52}{53}+\frac{52}{53}.\frac{4}{21}\)
=\(\frac{52}{53}.\left(\frac{17}{21}+\frac{4}{21}\right)\)
= \(\frac{52}{53}\)
`47/53*(17/3-53/47)+17/3*(6/17-47/53)`
`=47/53*17/3-1+6/3-17/3*47/53`
`=2-1+17/3*47/53-17/3*47/53`
`=1`
\(\dfrac{47}{53}.\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}.\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\)
\(=\left(\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{17}{3}.\dfrac{47}{53}\right)+\left(\dfrac{-47}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}\right)\)
\(=\left(1-1\right)+\left(-1+2\right)\)
\(=0+1\)
\(=1\)
53 x 39 + 47 x 39 - 53 x 21 - 47 x 21
= 53 ( 39 - 21 ) + 47 ( 39 - 21 )
= 53 x 18 + 47 x 18
= ( 53 + 47 ) x 18
= 100 x 18
= 1 800
Có:
\(\frac{5^3}{8.13}+\frac{5^3}{13.18}+...+\frac{5^3}{93.98}\)
= \(5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{98}\right)\)
=\(\frac{1125}{392}\)
=> \(\frac{1125}{392}.3\frac{17}{125}\)
= ...
cảm ơn bạn nhá!