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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, \(\frac{3.7.13.37.39-10101}{505050-70707}=\frac{10101.39-10101}{50.10101-7.10101}\)
\(=\frac{10101.\left(39-1\right)}{10101.\left(50-7\right)}\)
\(=\frac{10101.38}{10101.43}\)
\(=\frac{38}{43}\)
b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)
\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)
\(=\dfrac{2}{7}\)
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 = \(\frac{-3}{8}\)
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,\(\frac{2929-101}{3838+404}\)\(=\frac{2828}{4242}=\frac{2}{3}\)
\(b,\frac{54-34}{189-119}=\frac{20}{70}=\frac{2}{7}\)
\(c,d,e,f,f,g,h\)\(tuong\) \(tu\)
\(\frac{13.\left(2929-101\right)}{2.1919+404}=\frac{13\left(101.29-101\right)}{2.101.19+101.4}=\frac{13.101\left(29-1\right)}{2.101\left(19+2\right)}=\frac{13.28}{2.21}=\frac{13.2}{3}=\frac{26}{3}\)
