\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)*\(\frac{5}{5}\)
NH
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\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{\frac{-5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
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Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
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\(\frac{\frac{4}{2010}+\frac{4}{2011}-\frac{4}{2012}}{\frac{5}{2010}+\frac{5}{2011}-\frac{5}{2012}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{37}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{37}+1}\)= ?
Tính
\(P = 2018.(\frac{\frac{1}{3} - \frac{1}{5} + \frac{1}{7}}{\frac{5}{3} - 1 +\frac{5}{7}} + \frac{1 +\frac{4}{5} -\frac{2}{3}}{\frac{5}{4} + 1 -\frac{5}{6}}) : \frac{20182018}{20192019}\)
Bạn lấy 1/5 ở cả phân số 1 và 2 làm thừa số chung sau đó rút gọn và sẽ tìm đc kết qyar là 0
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\(P=2018.\left(\frac{\frac{1}{3}-\frac{1}{5}+\frac{1}{7}}{\frac{5}{3}-1+\frac{5}{7}}+\frac{1+\frac{4}{5}-\frac{2}{3}}{\frac{5}{4}+1-\frac{5}{6}}\right):\frac{20182018}{20192019}\)
\(P=\frac{\frac{1}{3}-\frac{1}{5}+\frac{1}{7}}{\frac{5}{3}-1+\frac{5}{7}}+\frac{1+\frac{4}{5}-\frac{2}{3}}{\frac{5}{4}+1-\frac{5}{6}}:\frac{20182018}{20192019}\)
\(P=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{5}}{\frac{5}{3}+\frac{5}{7}-1}+\frac{1+\frac{4}{5}-\frac{2}{3}}{1+\frac{5}{4}-\frac{5}{6}}:\frac{20182018}{20192019}\)
\(P=20192019\left(\frac{1+\frac{4}{5}-\frac{2}{3}}{1+\frac{5}{4}-\frac{5}{6}}+\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{5}}{\frac{5}{3}+\frac{5}{7}-1}\right):20182018\)
\(P=2019\left(\frac{1+\frac{4}{5}-\frac{2}{3}}{1+\frac{5}{4}-\frac{5}{6}}+\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{5}}{\frac{5}{3}+\frac{5}{7}-1}\right).2018\)
\(P=2019\left(\frac{1}{5}+\frac{4}{5}\right):2018\)
\(P=2019.1:2018\)
\(P=\frac{2019}{2018}\)
\(P=2018.\frac{2019}{2018}\)
\(P=2019\)
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\(\frac{\frac{1}{3}-\frac{5}{2}}{\frac{3}{4}-\frac{1}{2}}\cdot\frac{\frac{5}{6}+\frac{7}{3}}{1-\frac{5}{6}}\cdot\frac{-\frac{2}{5}+1}{\frac{2}{5}-1}\)
Afrac{frac{1}{3}-frac{5}{2}}{frac{3}{4}-frac{1}{2}}.frac{frac{5}{6}+frac{7}{3}}{1-frac{5}{6}}.frac{-frac{2}{5}+1}{frac{2}{5}+1}
Bfrac{frac{1}{3}-frac{4}{5}}{frac{1}{3}-frac{4}{5}}.frac{frac{3}{4}-frac{5}{3}}{frac{3}{4}+frac{5}{3}}:frac{frac{4}{5}-1}{1-frac{2}{3}}
mong mọi người giải giúp em
Đọc tiếp
A=\(\frac{\frac{1}{3}-\frac{5}{2}}{\frac{3}{4}-\frac{1}{2}}\).\(\frac{\frac{5}{6}+\frac{7}{3}}{1-\frac{5}{6}}\).\(\frac{-\frac{2}{5}+1}{\frac{2}{5}+1}\)
B=\(\frac{\frac{1}{3}-\frac{4}{5}}{\frac{1}{3}-\frac{4}{5}}\).\(\frac{\frac{3}{4}-\frac{5}{3}}{\frac{3}{4}+\frac{5}{3}}\):\(\frac{\frac{4}{5}-1}{1-\frac{2}{3}}\)
mong mọi người giải giúp em
Cho \(S_1=1+\frac{1}{5}\), \(S_2=1+\frac{1}{5}+\frac{1}{5^2}\), \(S_3=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\), tới \(S_n=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...........+\frac{1}{5^n}\). Chứng minh rằng : \(A=\frac{1}{5S_1^2}+\frac{1}{5^2S_2^2}+\frac{1}{5^3S_3^2}+\frac{1}{5^4S_4^2}+..........+\frac{1}{5^nS_n^2}<\frac{35}{36}\)
Khi \(n=1\to A=\frac{1}{5S_1^2}=\frac{5}{36}S_{k-1}\to S^2_k>S_k\cdot S_{k-1}\).
Vậy ta có \(\frac{1}{5^kS_k^2}
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\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\)
\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}\)
\(\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)
\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)
\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)
\(=\frac{5}{7}\times\frac{-5}{12}\)
\(=\frac{5\times\left(-5\right)}{7\times12}\)
\(=\frac{-25}{84}\)
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\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)
\(=\frac{5}{7}.\frac{-5}{12}\)
\(=-\frac{25}{84}\)
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\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\frac{1}{4}\)
\(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{5}{12}-\frac{1}{4}\)
\(\frac{3}{4}-\frac{3}{2}+\frac{5}{24}-\frac{1}{4}\)
\(\frac{18}{24}-\frac{36}{24}+\frac{5}{24}-\frac{6}{24}\)
-19/24
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Xem thêm câu trả lời
chung minh rang:
\(S=\frac{1}{5^2}-\frac{1}{5^4}+\frac{1}{5^6}-...+\frac{1}{5^{4n-2}}-\frac{1}{5^{4n}}+...+\frac{1}{5^{2010}}-\frac{1}{5^{^{2012}}}
\(A=\frac{\frac{1}{3}-\frac{5}{2}.\frac{5}{6}+\frac{7}{3}.-\frac{2}{5}+1}{\frac{3}{4}-\frac{1}{2}.1-\frac{5}{6}.\frac{2}{5}-1}\)
Tính
C= \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{2\frac{2}{3}-1+\frac{5}{7}}+\frac{1+\frac{4}{5}-\frac{2}{3}}{\frac{5}{4}+1-\frac{5}{6}}\)