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

Công tử họ Nguyễn cấm sủa

=-1

=2/5

=38/43

= 3/2

a)\(\dfrac{3.7.13.37.39-10101}{505050-70707}\)=\(\dfrac{10101.39-10101}{10101.50-10101.7}\)=\(\dfrac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)=\(\dfrac{39-1}{50-7}\)=\(\dfrac{38}{43}\)

b)\(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)=\(\dfrac{36.17+36.\left(-62\right)}{36.\left(-17\right)+36.\left(-13\right)}\)=

\(\dfrac{36.\left[17+\left(-62\right)\right]}{36.\left[\left(-17\right)+\left(-13\right)\right]}\)=\(\dfrac{36.\left(-45\right)}{36.\left(-20\right)}\)=\(\dfrac{45}{20}\)=\(\dfrac{9}{4}\)=2\(\dfrac{1}{4}\)

dễ mà

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) \(\dfrac{18.34-18.124}{-36.17+9.\left(-4\right).13}=\dfrac{18.\left(34-124\right)}{-36.\left(17+13\right)}=\dfrac{18.\left(-90\right)}{36.\left(-30\right)}=\dfrac{1.\left(-3\right)}{2.\cdot\left(-1\right)}=\dfrac{3}{2}\)

b) \(\dfrac{393939-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101.\left(39-1\right)}{10101.\left(50-7\right)}=\dfrac{38}{43}\)

a) \(\dfrac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}=\dfrac{36.17+36.\left(-62\right)}{36.\left(-17\right)+35.\left(-13\right)}=\dfrac{36\left[17+\left(-62\right)\right]}{36\left[\left(-17\right)+\left(-13\right)\right]}=\dfrac{45}{20}=\dfrac{9}{4}\)

b) \(\dfrac{3.7.13.37.39-10101}{505050-70707}=\dfrac{10101.39-10101}{10101.50-10101.7}=\dfrac{10101\left(39-1\right)}{10101\left(50-7\right)}=\dfrac{38}{43}\)

Phân số a quá dễ, bạn chỉ việc giản ước tử với mẫu sẽ được \(\frac{1}{-1995}\)

b) \(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)

Sau khi giản ước ta có: 

\(\frac{2+2+2}{5+5+5}=\frac{8}{15}\)

\(\frac{-1997.1996+1}{\left(1996-1\right).1997+1996}=\frac{-1997.1996+1}{1996.1997-1997+1996}=\frac{-\left(1997.1996-1\right)}{1996.1997-1}=-1\)

\(\frac{2.3.\left(1+2.3+7.7\right)}{3.5.\left(1+3.2+7.7\right)}=\frac{2}{5}\)

\(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{10101.50-10101.7}=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}=\frac{38}{43}\)

\(\frac{-5^3.2.2^6}{3^3.5.2^{14}.5}=\frac{-5^3.2^7}{3^3.5^2.2^{14}}=\frac{-5}{3^3.2^7}\)