Tính nhanh BT sau:
\(\dfrac{1}{500}+\dfrac{3}{500}+\dfrac{5}{500}+......+\dfrac{95}{500}+\dfrac{97}{500}+\dfrac{99}{500}\)
Những câu hỏi liên quan
1) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}< \dfrac{5}{16}\)
1)
\(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\\ 5A=1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\\ 5A-A=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\right)\\ 4A=1-\dfrac{500}{5^{500}}\\ A=\left(1-\dfrac{500}{5^{500}}\right):4\\ A=1:4-\dfrac{500}{5^{500}}:4\\ A=\dfrac{1}{4}-\dfrac{500}{5^{500}\cdot4}< \dfrac{1}{4}< \dfrac{5}{16}\)
Vậy \(A< \dfrac{5}{16}\)
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9 tháng 5 2021 lúc 20:08
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
Mình làm được một câu thôi, bạn dựa vào làm nha!
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\(\dfrac{24.135+3.561.8+4.126.6}{1+3+5+7+...+97+99-500}\)
Đặt \(A=\dfrac{24\cdot135+3\cdot561\cdot8+4\cdot126\cdot6}{1+3+5+7+...+97+99-500}\)
\(=\dfrac{24\cdot822}{2000}=\dfrac{1233}{125}\)
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tính H = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\)
Tính \(H=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.............+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-..............\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+........+\dfrac{1}{500}}\)
Help me!!!
Đặt vế đầu là A, vế sau là B.
Vế A:
- Tử:
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Vậy:
\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)
Vế B:
- Tử:
\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)
Vậy:
\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)
Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)
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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
tính:
\(2\sqrt{5}\)+\(\dfrac{3}{4}\sqrt{80}\)-0,3\(\sqrt{500}\)-\(\dfrac{1}{5}\sqrt{125}\)
\(=2\sqrt{5}+3\sqrt{5}-3\sqrt{5}-\sqrt{5}=\sqrt{5}\)
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Chứng minh rằng:
1) B dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{19}1
2) Adfrac{1}{5}+dfrac{2}{5^2}+dfrac{3}{5^3}+dfrac{4}{5^4}+...+dfrac{500}{5^{500}}100
3) Cdfrac{1}{2^3}+dfrac{1}{3^3}+dfrac{1}{4^3}+dfrac{1}{5^3}+...+dfrac{1}{500^3}dfrac{1}{4}
4) Ddfrac{4}{3}+dfrac{10}{9}+dfrac{28}{27}+...+dfrac{3^{98}+1}{3^{98}}100
Làm giúp mình sớm nha! Thanks.
Đọc tiếp
Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.
500+40+\(\dfrac{5}{10}\) +\(\dfrac{3}{1000}\) =?
=500 + 40 + 0,5 +0,003
=540 + 0,5 + 0,003
= 540,503
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