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\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)
Do đó:
\(M=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)
\(=\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}.\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}.\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}...\dfrac{\left(30-1\right)\left(30+1\right)}{30^2}\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{29.31}{30^2}=\dfrac{1.2.3...29}{2.3.4...30}.\dfrac{3.4.5...31}{2.3.4...30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
\(M=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{900}\right)\)
\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)
\(=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{3^2-1}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{30^2-1}{30^2}\right)\)
\(=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{29.31}{30^2}\right)\)
\(=\left(\dfrac{1.2.3...29}{2.3.4...30}\right).\left(\dfrac{3.4.5...31}{2.3.4...30}\right)=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
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