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\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}\)
\(x=\frac{23}{11}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
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\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)
\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)
\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)
\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)
\(B=\frac{100\cdot2}{1\cdot101}\)
\(B=\frac{200}{101}\)
Tính tổng dãy dấu ngoặc trước
Đặt \(S=1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot(4-1)+...+98\cdot99\cdot(100-97)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot3\cdot4+...+98\cdot99\cdot100-97\cdot98\cdot99\)
\(3S=98\cdot99\cdot100\Rightarrow S=\frac{1}{3}\cdot98\cdot99\cdot100\)
Thay vào đề bài,ta có :
\(\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}:\frac{-3}{2}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{90}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-30}{7}\cdot\frac{2}{-1}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-60}{-7}=\frac{60}{7}\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=\frac{60}{7}\cdot26950\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=231000\)
\(\Leftrightarrow323400\cdot x=231000\)
\(\Leftrightarrow x=231000:323400=\frac{5}{7}\)
Tử thần sai từ dòng:
\(\frac{\frac{1}{3}.98.99.100.x}{26950}=\frac{30}{7}.\frac{2}{-1}\Leftrightarrow12x=-\frac{60}{7}\Leftrightarrow x=\frac{-5}{7}\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
\(c,1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
\(d,\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=2\)
\(\left(1\cdot2\right)^{-1}+\left(2\cdot3\right)^{-1}+\cdot\cdot\cdot+\left(9\cdot10\right)^{-1}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)