a: \(\dfrac{-315}{540}=\dfrac{-315:45}{540:45}=\dfrac{-7}{12}\)

b: \(\dfrac{25\cdot13}{26\cdot35}=\dfrac{25}{35}\cdot\dfrac{1}{2}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

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, \(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{50.10101-7.10101}\)

                                                   \(=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)

                                                   \(=\frac{10101.38}{10101.43}\)

                                                   \(=\frac{38}{43}\)

mi chưa mần à?

Mk vt kq luôn ha bn!!

a) \(\dfrac{-315}{540}\)= \(\dfrac{-7}{12}\)

b) \(\dfrac{25.13}{26.35}\)= \(\dfrac{25.1}{2.35}\)= \(\dfrac{5}{14}\)

c) \(\dfrac{2929-101}{2.1919+404}\)= \(\dfrac{2828}{4242}\)

dễ quá đi. Nếu muốn mình giải hãy k trước đi

a) \(\frac{2929-101}{2.1919+404}\)\(=\frac{101.29-101}{2.1919+2.202}\)\(=\frac{101\left(29-1\right)}{2\left(1919+202\right)}\)\(=\frac{101.28}{2.2121}\)\(=\frac{2.101.7.2}{2.101.7.3}\)\(=\frac{2}{3}\)
b) \(\frac{-1997.1996+1}{-1995\left(-1997\right)+1996}\)\(=\frac{-1997.\left(1995+1\right)+1}{-1995\left(-1997\right)+1996}\)\(=\frac{-1997.1995-1997+1}{-1\left(-1997\right).1995+1996}\)\(=\frac{-1997.1995-1996}{1997.1995+1996}\)\(=\frac{-\left(1997.1995+1996\right)}{1997.1995+1996}\)\(=-1\)
c) \(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)\(=\frac{2.3+4.2.3+2.7.3.7}{3.5+2.3.2.5+3.7.5.7}\)\(=\frac{2.3\left(1+4+7.7\right)}{3.5\left(1+2.2+7.7\right)}\)\(=\frac{2\left(1+4+49\right)}{5\left(1+4+49\right)}\)\(=\frac{2}{5}\)
d) \(\frac{3.7.13.37.39-10101}{505050-70707}\)\(=\frac{3.7.13.37.39-3.7.13.37}{2.3.5.5.7.13.37-3.7.7.13.37}\)\(=\frac{3.7.13.37\left(39-1\right)}{3.7.13.37\left(2.5.5-7\right)}\)\(=\frac{38}{43}\)
e) \(\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)\(=\frac{18.2.17-18.2.62}{\left(-1\right).2.2.3.3.17+\left(-1\right)2.2.3.3.13}\)\(=\frac{18.2\left(17-62\right)}{\left(-1\right).2.2.3.3\left(17+13\right)}\)\(=\frac{\left(-1\right).2.2.3.3.45}{\left(-1\right).2.2.3.3.30}\)\(=\frac{45}{30}\)\(=\frac{3}{2}\)

a, \(\frac{29.101-101}{2.101.19+4.101}=\frac{101.\left(29-1\right)}{38.101+4.101}=\frac{101.28}{101.42}=\frac{28}{42}=\)\(\frac{2}{3}\)

b, \(\frac{-1997.1996+1}{-1995.\left(-1997\right)+1996}=\frac{-1997.\left(1995+1\right)+1}{1995.1997+1996}\)\(=\frac{-1997.1995+1997-1}{1995.1997+1996}=\frac{-1997.1995+1996}{1995.1997+1996}=-1\)

c,\(\frac{2.3+2\left(2.3\right)+7\left(2.3\right)}{3.5+2\left(3.5\right)+7\left(3.5\right)}=\frac{2.3\left(1+2+7\right)}{3.5\left(1+2+7\right)}=\frac{2.3}{3.5}=\frac{2}{5}\)

d,\(\frac{3.7.13.37.39-3.7.13.37}{2.3.5^2.7.13.37-3.7^2.13.37}=\frac{3.7.13.37.38}{3.7.13.37\left(2.5^2-7\right)}\)\(=\frac{3.7.13.37.38}{3.7.13.37.43}=\frac{38}{43}\)

e, \(\frac{18.34+18.\left(-124\right)}{9.\left(-4\right).17+9.\left(-52\right)}=\frac{18\left(-124+34\right)}{9.\left(-68\right)+9\left(-52\right)}\)\(=\frac{18\left(-90\right)}{9\left(-120\right)}=\frac{18.\left(-90\right)}{9.2\left(-60\right)}=\frac{18.\left(-90\right)}{18.\left(-60\right)}=\)\(\frac{-90}{-60}=\frac{3}{2}\)

d)\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+2.2.6+2.7.21}{3.5+3.2.10+3.7.35}=\frac{2.3+2.12+2.147}{3.5+3.20+3.245}=\frac{2\left(3+12+147\right)}{3\left(5+20+245\right)}\)

\(=\frac{2.162}{3.270}=\frac{54}{135}=\frac{2}{5}\)

\(a.\frac{-2019.2018+1}{\left(-2017\right).\left(-2019\right)+2018}\)

\(=\frac{2019.\left(-2018\right)+1}{2019.2017+2018}\)

\(=\frac{2019.\left(-2018\right)+1}{2019.2018-1}\)

\(=-\frac{2018}{2018}\)

\(=-1\)

c)\(\frac{2929-101}{2.1919+404}=\frac{29.101-101}{29.101+101.4}=\frac{101\left(29-1\right)}{101\left(29+4\right)}=\frac{28}{33}\)