rut gon phan so sau
\(\frac{3.13-13.18}{15.40-80}\)
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\(\frac{3\cdot13-13\cdot18}{15\cdot40-80}=\frac{13\left(3-18\right)}{15\cdot40-40\cdot2}=\frac{-15\cdot13}{40\cdot13}=-\frac{3}{8}\)
Sao ảnh hai bn Linh Le và Trần Việt Linh giống nhau thế hay là cùng nick
a, \(-\dfrac{315}{540}\) = \(\dfrac{-315:45}{540:45}\) = \(\dfrac{-7}{12}\) b, \(\dfrac{25.13}{26.35}\) = \(\dfrac{25.13:5:13}{26.35:13:5}\) = \(\dfrac{5}{14}\)
c, \(\dfrac{6.9-2.17}{63.3-119}\) = \(\dfrac{2.3.9-2.17}{7.9.3-7.17}\) = \(\dfrac{2.(27-17)}{7.(7-17)}\) = \(\dfrac{2}{7}\)
d, \(\dfrac{3.13-13.18}{15.40-80}\) = \(\dfrac{3.13(1-6)}{40.(15-2)}\) = \(\dfrac{-3.13.5}{40.13}\) = \(\dfrac{-15}{40}\) = \(\dfrac{-15:5}{40:5}\) = \(-\dfrac{3}{8}\)
Đặt A = \(\frac{3.13-13.18}{15.40-80}\)
=> A = \(\frac{13.\left(3-18\right)}{15.40-80}\)
=> A = \(\frac{13.\left(-15\right)}{15.40-40.2}\)
=> A = \(\frac{13.\left(-15\right)}{40.\left(15-2\right)}\)
=> A = \(\frac{13.\left(-15\right)}{40.13}\)
=> A = \(\frac{-15}{40}\)
=> A = \(\frac{-3}{8}\)
Rút gọn các phân số sau:
a, \(\dfrac{-315}{540}\) = \(\dfrac{-315:45}{540:45}\) = \(\dfrac{-7}{12}\)
b, \(\dfrac{25.13}{26.35}\) = \(\dfrac{5.5.13}{13.2.5.7}\) = \(\dfrac{5}{2.7}\) = \(\dfrac{5}{14}\)
c, \(\dfrac{3.13-13.18}{15.40-80}\) = \(\dfrac{3.13-13.18}{15.40-2.40}\) = \(\dfrac{13.\left(3-18\right)}{40.\left(15-2\right)}\)
= \(\dfrac{13.\left(-15\right)}{40.13}\) = \(\dfrac{-3}{8}\)
a) \(\dfrac{-315}{540}=\dfrac{-315:45}{540:45}=\dfrac{-7}{12}\)
b) \(\dfrac{25\cdot13}{26\cdot35}=\dfrac{5\cdot5\cdot13}{2\cdot13\cdot5\cdot7}=\dfrac{5}{2\cdot7}=\dfrac{5}{14}\)
c) \(\dfrac{3\cdot13-13\cdot18}{15\cdot40-80}=\dfrac{3\cdot13-13\cdot18}{15\cdot40-2\cdot40}=\dfrac{13\cdot\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13\cdot\left(-15\right)}{40\cdot13}=\dfrac{1\cdot\left(-3\right)}{8\cdot1}=\dfrac{-3}{8}\)
Trả lời:
a, \(\frac{6\times9-2\times17}{63\times3-119}=\frac{2.3\times9-2\times17}{7.9\times3-7.17}\)
\(=\frac{2\times\left(3\times9-17\right)}{7\times\left(3\times9-17\right)}\)
\(=\frac{2}{7}\)
b, \(\frac{3\times13-13\times18}{15\times40-80}=\frac{13\times\left(3-18\right)}{40\times\left(15-2\right)}\)
\(=\frac{13\times-15}{40\times13}\)
\(=\frac{-3}{8}\)
c, \(\frac{-1997.1996+1}{\left(-1995\right).\left(-1997\right)+1996}=\frac{-1997.1996+1}{\left(1-1996\right).\left(-1997\right)+1996}\)
\(=\frac{-1997.1996+1}{-1997-1996.\left(-1997\right)+1996}\)
\(=\frac{-1997.1996+1}{-1996.\left(-1997\right)-1}\)
\(=\frac{-1997.1996+1}{-\left[1996.\left(-1997\right)+1\right]}\)
\(=-1\)
d, \(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{50.10101-7.10101}\)
\(=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)
\(=\frac{10101.38}{10101.43}\)
\(=\frac{38}{43}\)
a)\(\dfrac{6.9-2\cdot17}{63\cdot3-119}=\dfrac{2\cdot3\cdot9-2\cdot17}{7\cdot9\cdot3-7\cdot17}=\dfrac{2\cdot27-2\cdot17}{7\cdot27-7\cdot17}=\dfrac{2\left(27-17\right)}{7\left(27-17\right)}=\dfrac{2}{7}\)
b) \(\dfrac{3\cdot13-13\cdot18}{15\cdot40-80}=\dfrac{13\cdot\left(3-18\right)}{\left(15-2\right)\cdot40}=\dfrac{13\cdot\left(-15\right)}{13\cdot40}=-\dfrac{15}{40}=-\dfrac{3}{8}\)
c) \(\dfrac{25\cdot13}{26\cdot35}=\dfrac{5^2\cdot13}{2\cdot13\cdot5\cdot7}=\dfrac{5}{14}\)
a) \(\dfrac{6.9-2.17}{63.3-119}=\dfrac{2.3.9-2.17}{7.9.3-17.7}=\dfrac{2.\left(3.9-17\right)}{7.\left(9.3-17\right)}=\dfrac{2}{7}\)
b)\(\dfrac{3.13-13.18}{15.40-80}=\dfrac{3.13-13.18}{15.40-40.2}=\dfrac{13.\left(3-18\right)}{40.\left(15-2\right)}=\dfrac{13.\left(-15\right)}{40.13}=\dfrac{-15}{40}=\dfrac{-3}{8}\)
c) \(\dfrac{25.13}{26.35}=\dfrac{5.5.13}{13.2.5.7}=\dfrac{5}{2.7}=\dfrac{5}{14}\)
\(\frac{3.13-13.18}{15.40-80}=\frac{13.\left(3-18\right)}{15.40-40.2}\)
\(=\frac{13.\left(-15\right)}{40.\left(15-2\right)}=\frac{13.\left(-15\right)}{40.13}\)
\(=\frac{-15}{13}\)
\(\frac{3.13-13.18}{15.40-80}\) \(=\frac{13.\left(3-18\right)}{40.\left(15-2\right)}=\frac{13.\left(-15\right)}{40.13}\) \(=\) \(\frac{-8}{3}\)