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a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)
\(=\frac{100}{2}=50\)
\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)
\(=-\frac{1}{2}.-\frac{2}{3}..........-\frac{99}{100}\)
\(=\frac{-1}{100}\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5.....100}{2.3.4.....99}\)
\(=\frac{100}{2}=50\)
Ta có :
M = \(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{1+\left(\frac{1}{99}+1\right)+\left(\frac{2}{98}+1\right)+\left(\frac{3}{91}+1\right)+...+\left(\frac{98}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{\frac{100}{100}+\frac{100}{99}+\frac{100}{98}+\frac{100}{97}+...+\frac{100}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(\frac{100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
M = \(100\)
N = \(\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\left(1-\frac{1}{9}\right)+\left(1-\frac{2}{10}\right)+\left(1-\frac{3}{11}\right)+...+\left(1-\frac{92}{100}\right)}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{500}}\)
N = \(\frac{8.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{5}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}\)
N = \(40\)
\(\Rightarrow\)M : N = \(\frac{100}{40}\%=250\%\)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
Câu a :
A = 2\(\frac12\) : (- \(\frac12\))^2 - \(\frac{1}{-3}\).(\(\frac{1}{-2}\) - \(\) \(\frac43\) : \(-\frac89\))
A = \(\frac52\) : \(\frac14\) + \(\frac13\).(-\(\frac12\) + \(\frac43\times\frac98\))
A = \(\frac52\times4\) + \(\frac13\).(- \(\frac12+\frac32\))
A = \(10\) + 1/3
A = 30/3 + 1/3
A = 31/3
Câu b:
B = (3\(\frac{10}{99}\) + 4\(\frac{11}{99}\) - \(\frac{58}{229}\)).(\(\frac12\) - \(\frac43\) - \(\frac16\))
B = (7\(\frac{21}{99}\) - \(\frac{58}{229}\))(3/6 - 8/6 - 1/6)
B = (7\(\frac{7}{33}\) - \(\frac{58}{229}\)).(-1)
B = - (\(\frac{238}{33}-\frac{58}{229}\))
B = - (\(54502\)/7557 - 1914/7557)
B = - 52588/7557
a)
\(=\frac{3}{2}.\frac{4}{3}......\frac{100}{99}=\frac{100}{2}=50\)
b)
\(=\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}.....\frac{\left(-99\right)}{100}=\frac{-1}{100}\)