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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) \(\frac{18.34+\left(-18\right).124}{-36.17+9.\left(-52\right)}\)\(=\frac{18.2.17-18.2.62}{\left(-1\right).2.2.3.3.17+\left(-1\right)2.2.3.3.13}\)\(=\frac{18.2\left(17-62\right)}{\left(-1\right).2.2.3.3\left(17+13\right)}\)\(=\frac{\left(-1\right).2.2.3.3.45}{\left(-1\right).2.2.3.3.30}\)\(=\frac{45}{30}\)\(=\frac{3}{2}\)
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,\(\frac{2929-101}{3838+404}\)\(=\frac{2828}{4242}=\frac{2}{3}\)
\(b,\frac{54-34}{189-119}=\frac{20}{70}=\frac{2}{7}\)
\(c,d,e,f,f,g,h\)\(tuong\) \(tu\)
Giải:
a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\)
\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\)
\(=\dfrac{1}{2^6}=\dfrac{1}{64}\)
b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\)
\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\)
\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\)
\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\)
\(=2\)
c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\)
\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\)
\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\)
\(=\dfrac{2.3}{3.5}\)
\(=\dfrac{2}{5}\)
Chúc bạn học tốt!
