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Ta có: \(M\cdot N\)
\(=\dfrac{1}{3}\cdot\dfrac{5}{7}\cdot\dfrac{7}{9}\cdot\dfrac{13}{15}\cdot...\cdot\dfrac{37}{39}\cdot\dfrac{7}{5}\cdot\dfrac{11}{9}\cdot...\cdot\dfrac{39}{37}\)
\(=\dfrac{1}{3}\)
Ta có: M⋅N
=1/3⋅5/7⋅7/9⋅13/15⋅...⋅37/39⋅7/5⋅11/9⋅...⋅39/37
=1/3
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\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}+\frac{1}{11x13}\)
\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}x\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}x\frac{10}{39}\)
\(=\frac{5}{39}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+\frac{2}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+\frac{11-9}{9\times11}+\frac{13-11}{11\times13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{13}\right)=\frac{6}{13}\)
Do đó ta có:
\(\frac{6}{13}\times y=\frac{3}{5}\)
\(\Leftrightarrow y=\frac{13}{10}\).
Ta có : \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{11.13}\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}\)
\(=\frac{10}{39}\)
\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{11\times13}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{11\times13}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
Nhưng bn ơiX là x^2 hay tách biệt nếu tách biệt thì là 9/49 còn nếu là x^2 thì là 3/7 nhé
\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)xX=\frac{9}{7} \)\(=\left(\frac{1}{3}-\frac{1}{21}\right)xX=\frac{9}{7}\)\(=\frac{2}{7}xX=\frac{9}{7}\)
\(X=\frac{9}{7}:\frac{2}{7}\)
\(X=\frac{9}{2}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\)\(...+\frac{2}{8.9}+\frac{2}{9.10}\)
Đặt \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}+\frac{2}{9.10}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}\)
\(A=\frac{4}{15}\)
\(B=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}+\frac{2}{9.10}\)
\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2\left(1-\frac{1}{10}\right)\)
\(B=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
\(\Rightarrow A+B=\frac{4}{15}+\frac{9}{5}\)
\(=\frac{31}{15}\)
Vậy biểu thức trên có giá trị là \(\frac{31}{15}\)
=2/5-2/7+ 2/7-2/9+2/9-2/11+2/11-2/13+2/13-2/15
=2/5-(2/7-2/7)-(2/9-2/9)-(2/11-2/11)-(2/13-2/13)-2/15
=2/5-0-0-0-0-2/15
=2/5-2/15
4/15