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

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

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

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

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: 

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

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

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

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

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

\(a,\dfrac{2929-101}{2.1919+404}=\dfrac{29.101-101.1}{2.19.101+4.101}\)

\(=\dfrac{101\left(29-1\right)}{101\left(2.19+4\right)}\)

\(=\dfrac{101.29}{101.42}\)

\(=\dfrac{28}{42}=\dfrac{2}{3}\)

\(b,\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}.\left(-100\right)^0}=\dfrac{\left(-5\right)^3.40.4^3}{135.\left(-2\right)^{14}}\)

\(=\dfrac{40.\left(-15\right).64}{135.\left(-2\right)^{14}}\)

\(=\dfrac{5.8.3.\left(-5\right).64}{5.3.9.\left(-2\right)^{14}}\)

\(=\dfrac{8.\left(-5\right).\left(-2\right)^6}{9.\left(-2\right)^{14}}\)

\(=\dfrac{\left(-2\right)^3.5}{9.\left(-2\right)^8}=\dfrac{5}{9.\left(-2\right)^5}\)

Bài 1:

Coi \(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\)

\(2A=\left(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\right).2\)

\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)

\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=5.\left(1-\frac{1}{101}\right)\)

\(=5.\frac{100}{101}\)

\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)

\(\left(\frac{5}{1}-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}....+\frac{5}{99}-\frac{5}{101}\right):\frac{1}{5}\)

\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}.....+\frac{1}{99}-\frac{1}{101}\right).5\)

\(\left(\frac{1}{1}-\frac{1}{101}\right).5\)

\(\frac{100}{101}.5\)

\(\frac{500}{101}\)

2,a,\(\frac{2929-101}{3838+404}\)\(=\frac{2828}{4242}=\frac{2}{3}\)

\(b,\frac{54-34}{189-119}=\frac{20}{70}=\frac{2}{7}\)

\(c,d,e,f,f,g,h\)\(tuong\) \(tu\)

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...............+\frac{5}{99.101}\)

\(=\frac{5.2}{1.3.2}+\frac{5.2}{3.5.2}+\frac{5.2}{5.7.2}+............+\frac{5.2}{99.101.2}\)

\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+.................+\frac{2}{99.101}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\frac{100}{101}\)

\(=\frac{250}{101}\)