Violympic toán 6
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rút gọn
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}}{\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}}\)
20 tháng 3 2021 lúc 16:09
Giúp mk vớiCâu 1:Cho A dfrac{1}{dfrac{99}{dfrac{1}{2}+}}+dfrac{2}{dfrac{98}{dfrac{1}{3}+}}+dfrac{3}{dfrac{97}{dfrac{1}{4}+....}}+...+dfrac{99}{dfrac{1}{dfrac{1}{100}}}. B dfrac{92}{dfrac{1}{45}+}-dfrac{1}{dfrac{9}{dfrac{1}{50}+}}-dfrac{2}{dfrac{10}{dfrac{1}{55}+}}-dfrac{3}{dfrac{11}{dfrac{1}{60}+....}}-...dfrac{92}{dfrac{100}{dfrac{1}{500}}}. Tính dfrac{A}{B}
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Giúp mk với
Câu 1:
Cho A = \(\dfrac{1}{\dfrac{99}{\dfrac{1}{2}+}}+\dfrac{2}{\dfrac{98}{\dfrac{1}{3}+}}+\dfrac{3}{\dfrac{97}{\dfrac{1}{4}+....}}+...+\dfrac{99}{\dfrac{1}{\dfrac{1}{100}}}\).
B =\(\dfrac{92}{\dfrac{1}{45}+}-\dfrac{1}{\dfrac{9}{\dfrac{1}{50}+}}-\dfrac{2}{\dfrac{10}{\dfrac{1}{55}+}}-\dfrac{3}{\dfrac{11}{\dfrac{1}{60}+....}}-...\dfrac{92}{\dfrac{100}{\dfrac{1}{500}}}\). Tính \(\dfrac{A}{B}\)
So sánh A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}\) và B = \(\dfrac{304}{1975}\)
so sanh
M=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)va \(\dfrac{1}{2}\)
B=\(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}va\dfrac{9}{10}\)
C=\(\dfrac{10}{17}+\dfrac{8}{15}+\dfrac{11}{16}va2\)
CMR\(\dfrac{1}{5}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}< \dfrac{2}{5}\)
a/ Rút gọn 2 biểu thức sau: \(E=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\)và \(F=\dfrac{94-\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}}{\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}}\)
b/ Tính E - 2F
cho A=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
CMR:0,2<A<0,4
A=(1+1/2).(1+1/3).(1+1/4)...(1×1/2009)
B=(1-1/2).(1-1/3)...(1-1/100)
B= 1/2.2/3.3/4...99/100
X+1/99+x+2/98+x+3/97+x+4/96
TÍNH GIÁ TRỊ BIỂU THỨC:\(\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{4}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot...\cdot\left(1+\dfrac{2}{97}\right)\cdot\left(1+\dfrac{2}{98}\right)\)