\(A=\frac{1+3+5+...+19}{21+23+25+..+39}\)
\(B=\frac{1.3.5...39}{21.22.23...40}\)
\(C=\frac{\left(6\frac{1}{2}-8\right):0,05}{\left(7\frac{1}{20}-5,65\right).6+1\frac{3}{5}}\)
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Những câu hỏi liên quan
Tính M=\(\frac{\left(6\frac{1}{2}-8\right):0,05}{\left(7\frac{1}{20}-5,65\right).6+1\frac{3}{5}}\)
Tính
a) \(\frac{\left(6\frac{1}{2}-8\right):0,05}{\left(7\frac{1}{20}-5,65\right).6+1\frac{3}{5}}\)
Tính M = \(\frac{\left(6\frac{1}{2}-8\right):0,05}{\left(7\frac{1}{20}-5,65\right).6+1\frac{3}{5}}\)
Lời giải:
\(M=\frac{\frac{-3}{2}:0,05}{1,4.6+1,6}=\frac{-30}{10}=-3\)
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Rút gọn biểu thức:
A = \(\frac{\left(6\frac{1}{2}-8\right):0,05}{\left(7\frac{1}{20}-5,65\right).6+1\frac{3}{5}}\)
Chứng minh rằng:
a)\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)với n thuộc N*
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
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Tính giá trị biểu thức1.Afrac{1}{5}+frac{3}{17}-frac{4}{3}+left(frac{4}{5}-frac{3}{17}+frac{1}{3}right)-frac{1}{7}+left[frac{-14}{30}right]2.Bleft(frac{5}{8}-frac{4}{12}+frac{3}{2}right)-left(frac{5}{8}+frac{9}{13}right)-left[frac{-3}{2}right]+frac{7}{-15}3.Cfrac{5}{18}+frac{8}{19}-frac{7}{21}+left(frac{-10}{36}+frac{11}{19}+frac{1}{3}right)-frac{5}{8}4.Dfrac{1}{9}-left[frac{-5}{23}right]-left(frac{-5}{23}+frac{1}{9}+frac{25}{7}right)+frac{50}{14}-frac{7}{30}5.Efrac{1}{13}+left(frac{-5}{18}-frac...
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Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
1)CMR:
a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b) \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)( n thuộc N* )
Chứng minh rằng:
a,\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b,\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)
Biết rằng n thuộc N*
a) Ta có:
\(\frac{1.3.5...39}{21.22.23...40}=\frac{1.3.5.7.11.13.15.17.19}{22.24.26.28.30.32.34.36.38}\)=\(\frac{1.3.5.7.9.11.13.15.17.19}{2.11.2^3.3.2.13.2^2.7.2.15.2^5.2.17.2^2.9.2.19.2^3.5}\)=\(\frac{1}{2.2^3.2.2^2.2.2^5.2.2^2.2.2^3}\)=\(\frac{1}{2^{1+3+1+2+1+5+1+2+1+3}}\)=\(\frac{1}{2^{20}}\)
Vậy \(\frac{1.3.5...39}{21.22.23...40}\)= \(\frac{1}{2^{20}}\)
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tick cho mk hết âm đi mk chân thành cẳm ơn
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Chứng minh rằng:
a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1}{2^n}\)
Nhân cả tử và mẫu vs:
a)2.4.6.....40
b)2.4.6.....2n
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