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`Answer:`
Ta thấy:
\(9=1.9\)
\(20=10.2\)
\(33=11.3\)
...
\(9200=100.92\)
`=>` Mẫu thức của từng nhân tử có dạng là \(n\left(n+8\right)\)
Xét dạng tổng quát của nhân tử: \(1+\frac{7}{n\left(n+8\right)}=\frac{n^2+8n+7}{n\left(n+8\right)}=\frac{\left(n+1\right)\left(n+7\right)}{n\left(n+8\right)}\)
\(n=1\Rightarrow1+\frac{7}{1.9}=\frac{2.8}{1.9}\)
\(n=2\Rightarrow1+\frac{7}{2.10}=\frac{3.9}{2.10}\)
\(n=3\Rightarrow1=\frac{7}{3.10}=\frac{4.10}{3.11}\)
...
\(n=92\Rightarrow1+\frac{7}{92.100}=\frac{93.99}{92.100}\)
\(\Rightarrow\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}...\frac{93.99}{92.100}=\frac{\left(2.3.4...93\right)\left(8.9.10...9\right)}{\left(1.2.3...92\right)\left(9.10.11...100\right)}=\frac{93.8}{1.100}=\frac{186}{25}\)
\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)\cdot\cdot\cdot\left(1-\frac{2}{2011}\right)\)
\(A=\left(\frac{5-2}{5}\right)\left(\frac{7-2}{7}\right)\left(\frac{9-2}{9}\right)\cdot\cdot\cdot\left(\frac{2011-2}{2011}\right)\)
\(A=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot\cdot\cdot\frac{2009}{2011}\)(các thừa số trên tử giống dưới mẫu mình lượt bỏ đi nhé!)
\(A=\frac{3}{2011}\)
\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{2011}\right)\)
\(=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{2009}{2011}\)
\(=\frac{3}{2011}\)
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=-2+3-\frac{31}{20}\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)
Học good
\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}\)
\(=-\frac{101}{200}\)
\(\left(\frac{3}{4}\right)^{18}.\left(\frac{-3}{4}\right)^5=\left(\frac{3}{4}\right)^{18+5}.\left(-1\right)=-\left(\frac{3}{4}\right)^{22}\)
KQ không hiển thị được phân số ...
\(\left(\frac{3}{4}\right)^{18}.\left(\frac{-3}{4}\right)^5=\left(\frac{3}{4}\right)^{18}.\left(-1\right)^5.\left(\frac{3}{4}\right)^5=\left(\frac{3}{4}\right)^{23}.\left(-1\right)\)
\(=-\left(\frac{3}{4}\right)^{23}\)