Tính nhanh:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
B = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)
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Tính nhanh:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
B = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)
Cứu mai thi rồi
Ta có :
M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)
b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)
Ta có:
B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)
Vậy B= \(\dfrac{100}{101}\)
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Bài 5: Tính nhanh tổng sau(nếu có):Mdfrac{3}{2}-dfrac{5}{6}+dfrac{7}{12}-dfrac{9}{20}+dfrac{11}{30}-dfrac{13}{42}+dfrac{15}{56}-dfrac{17}{72} ; Adfrac{5}{1.3}+dfrac{5}{3.5}+dfrac{5}{5.7}+.....+dfrac{5}{2019.2021}
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Bài 5: Tính nhanh tổng sau(nếu có):
M=\(\dfrac{3}{2}\)-\(\dfrac{5}{6}\)+\(\dfrac{7}{12}\)-\(\dfrac{9}{20}\)+\(\dfrac{11}{30}\)-\(\dfrac{13}{42}\)+\(\dfrac{15}{56}\)-\(\dfrac{17}{72}\) ; A=\(\dfrac{5}{1.3}\)+\(\dfrac{5}{3.5}\)+\(\dfrac{5}{5.7}\)+.....+\(\dfrac{5}{2019.2021}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
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1)Cho B=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\).Chứng minh B>1
2)Tính nhanh:M=\(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
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Adfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1
dfrac{2n+5}{n+2}
Sdfrac{1}{20}+dfrac{1}{21}+dfrac{1}{22}+L+dfrac{1}{29}
Adfrac{7}{4}.left(dfrac{3333}{1212}+dfrac{3333}{2020}+dfrac{3333}{3030}+dfrac{3333}{4242}right)
Adfrac{1}{4.5}+dfrac{1}{5.6}+...+dfrac{1}{48.49}+dfrac{1}{49.50}
Adfrac{17}{1.3}+dfrac{17}{3.5}+dfrac{17}{5.7}+...+dfrac{17}{49.51}
dfrac{2n+5}{3n+1}
left(left|xright|+dfrac{2}{5}right):dfrac{2}{5}1
dfrac{1}{2}+dfrac{1}{4}+dfrac{1}{8}+dfrac{1}{16}+dfra...
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\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(\dfrac{2n+5}{n+2}\)
\(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+L+\dfrac{1}{29}\)
\(A=\dfrac{7}{4}.\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}\)
\(A=\dfrac{17}{1.3}+\dfrac{17}{3.5}+\dfrac{17}{5.7}+...+\dfrac{17}{49.51}\)
\(\dfrac{2n+5}{3n+1}\)
\(\left(\left|x\right|+\dfrac{2}{5}\right):\dfrac{2}{5}=1\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}< 1\)
\(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(5.\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{10}\right)\le\dfrac{x}{20}\le-3\left(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}\right)\left(x\in Z\right)\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}< 1\)
trả hiểu yêu cầu đề bài là j cả
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1/.dfrac{1}{1}.dfrac{1}{2}+dfrac{1}{2}.dfrac{1}{3}+dfrac{1}{3}.dfrac{1}{4}+dfrac{1}{4}.dfrac{1}{5}
2/.dfrac{1}{2}+dfrac{1}{6}+dfrac{1}{12}+...+dfrac{1}{10100}
3/.A dfrac{2}{1.3}+dfrac{2}{3.5}+dfrac{2}{5.7}+...+dfrac{2}{99.101}
4/.A dfrac{1}{1.3}+dfrac{1}{3.5}+dfrac{1}{5.7}+...+dfrac{1}{99.101}
tính bằng cách thuận tiện nhất ( làm nhanh trước 5h nha , nếu ai làm được thì cho 100 tick , thật đó và trình bày cách diễn giải nha )
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1/.\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}\)
2/.\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{10100}\)
3/.A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
4/.A = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
tính bằng cách thuận tiện nhất ( làm nhanh trước 5h nha , nếu ai làm được thì cho 100 tick , thật đó và trình bày cách diễn giải nha )
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
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=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
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Tính nhanh:
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\)\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\)
giúp mình với
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
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1.Tính a,5text{x}dfrac{7}{3} b,dfrac{13}{4}:72.Tính a,dfrac{3}{7}+dfrac{2}{5}+dfrac{3}{4} b,dfrac{9}{7}-dfrac{5}{11}text{x}dfrac{11}{7} c,dfrac{3}{5}text{x}dfrac{5}{7}text{+}dfrac{4}{7} d,dfrac{7}{9}text{x}dfrac{2}{5}:dfrac{3}{11} e,dfrac{9}{7}+dfrac{2}{3}-dfrac{1}{4} g,dfrac{4}{9}:dfrac{3}{5}text{x}dfrac{2}{11} h,dfrac{7}{2}-dfrac{3}{10}:dfrac{2}{5}
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1.Tính
\(a,5\text{x}\dfrac{7}{3}\) \(b,\dfrac{13}{4}:7\)
2.Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\) \(b,\dfrac{9}{7}-\dfrac{5}{11}\text{x}\dfrac{11}{7}\) \(c,\dfrac{3}{5}\text{x}\dfrac{5}{7}\text{+}\dfrac{4}{7}\) \(d,\dfrac{7}{9}\text{x}\dfrac{2}{5}:\dfrac{3}{11}\) e,\(\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
g,\(\dfrac{4}{9}:\dfrac{3}{5}\text{x}\dfrac{2}{11}\) h,\(\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
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\(\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{60}{140}+\dfrac{56}{140}+\dfrac{105}{140}=\dfrac{221}{140}\)
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\(\dfrac{9}{7}-\dfrac{5}{11}x\dfrac{11}{7}=\dfrac{9}{7}-\dfrac{5}{7}=\dfrac{4}{7}\)
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3 tháng 3 2023 lúc 19:18
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
tính hợp lý:
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)
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`4/1.3+4/3.5+4/5.7+...+4/99.101`
`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`
`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`
`=2(1-1/101)`
`=2. 100/101`
`=200/101`
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17 tháng 3 2023 lúc 20:16
Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
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