\(\frac{3}{4}:5+\frac{5}{8}x2\)
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Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)
\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)
Vậy \(x_1=x_2=x_3=x_4=x_5=6\)
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\(1\frac{1}{3}\times2\frac{1}{4}\times...\times8\frac{1}{10}\)
\(=\frac{4}{3}\times\frac{9}{4}\times...\times\frac{81}{10}\)
\(=\frac{4\times9\times...\times81}{3\times4\times...\times10}\)
\(=\frac{2\times2\times3\times3\times...\times9\times9}{3\times4\times5\times...\times10}\)
\(=\frac{\left(2\times3\times4\times...\times9\right)\times\left(2\times3\times4\times...\times9\right)}{3\times4\times5\times...\times10}\)
\(=\frac{2\times2\times3\times...\times9}{10}\)
\(=2\times3\times4\times6\times...\times9\)
\(=72576\)
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\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
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B=3/2 xin loi nhavì cách trình bày trên này khó quá, đọc chắc bạn ko hiểu đâu
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Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
\(=\frac{3}{4}x\frac{1}{5}+\frac{5}{8}x2\)
\(=\frac{3}{20}+\frac{10}{8}\)
\(=\frac{6}{40}+\frac{50}{40}\)
\(=\frac{56}{40}\)
\(=\frac{7}{5}\)