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- \(\dfrac{13}{49}\) + \(\dfrac{12}{48}\) + \(\dfrac{1}{12}\) + \(\dfrac{3}{18}\)
= - \(\dfrac{13}{49}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{6}\)
= - \(\dfrac{13}{49}\) + ( \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\) + \(\dfrac{2}{12}\))
= - \(\dfrac{13}{49}\) + \(\dfrac{1}{2}\)
= - \(\dfrac{26}{98}\) + \(\dfrac{49}{98}\)
= \(\dfrac{23}{98}\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(\frac{-1}{4}-\frac{5}{3}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=\left(\frac{-3}{12}-\frac{20}{12}+\frac{15}{12}\right)+\left(\frac{7}{33}-\frac{18}{33}\right)+\frac{48}{49}\)
\(=\left(\frac{-23}{12}+\frac{15}{12}\right)+\left(\frac{-11}{33}\right)+\frac{48}{49}\)
\(=\frac{-2}{3}+\left(\frac{-1}{3}\right)+\frac{48}{49}\)
\(=-1+\frac{48}{49}\)
\(=-\frac{1}{49}\)
Ùng hộ mk nha ^_-
a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
=0
`(x-9)/48+(x+6)/12=(x-3)/4`
`=>x-9+4(x+6)=12(x-3)`
`=>x-9+4x+24=12x-36`
`=>5x+15=12x-36`
`=>7x=51`
`=>x=51/7`
Vậy `x=51/7`
Ta có: \(\dfrac{x-9}{48}+\dfrac{x+6}{12}=\dfrac{x-3}{4}\)
\(\Leftrightarrow\dfrac{x-9}{48}+\dfrac{4\left(x+6\right)}{48}=\dfrac{12\left(x-3\right)}{48}\)
\(\Leftrightarrow x-9+4x+24=12x-36\)
\(\Leftrightarrow5x+15-12x+36=0\)
\(\Leftrightarrow-7x+51=0\)
\(\Leftrightarrow-7x=-51\)
hay \(x=\dfrac{51}{7}\)
Vậy: \(x=\dfrac{51}{7}\)
a) \(=\left(\left(-\frac{1}{4}-\frac{5}{3}\right)+\frac{7}{33}\right)-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\left(-\frac{23}{12}+\frac{7}{33}\right)+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(-\frac{23}{12}+\frac{15}{12}\right)+\left(\frac{9}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=-\frac{2}{3}-\frac{3}{11}+\frac{48}{49}\)
\(=\frac{65}{1617}\)
b) \(=\frac{11}{125}+\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)