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\(\frac{1717}{1212}x+\frac{1717}{2020}x+\frac{1717}{3030}x=\frac{17}{6}\)
=>\(\left(\frac{1717}{1212}+\frac{1717}{2020}+\frac{1717}{3030}\right)x=\frac{17}{6}\)
=>\(\frac{17}{6}x=\frac{17}{6}\)
=>x=1
\(\frac{1717}{1212}x+\frac{1717}{2020}x+\frac{1717}{3030}x=\frac{17}{6}\Rightarrow\left(\frac{17}{12}+\frac{17}{20}+\frac{17}{30}\right)x=\frac{17}{6}\Rightarrow\frac{17}{6}x=\frac{17}{6}\Rightarrow x=1\)
Ta có: \(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{171}-\frac{3}{1717}}{\frac{9}{7}-\frac{9}{17}+\frac{9}{171}-\frac{9}{1717}}\)= \(\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{171}-\frac{1}{1717}\right)}{9.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{171}-\frac{1}{1717}\right)}\)=\(\frac{3}{9}=\frac{1}{3}\)
\(\left(\frac{151515}{161616}+\frac{17^9}{17^{10}}\right)-\left(\frac{1500}{1600}-\frac{1616}{1717}\right)\)
\(=\left(\frac{15}{16}+\frac{1}{17}\right)-\left(\frac{15}{16}-\frac{16}{17}\right)\)
\(=\frac{15}{16}+\frac{1}{17}-\frac{15}{16}+\frac{16}{17}\)
\(=\left(\frac{15}{16}-\frac{15}{16}\right)+\left(\frac{1}{17}+\frac{16}{17}\right)\)
\(=\frac{15-15}{16}+\frac{1+16}{17}\)
\(=0+\frac{17}{17}\)
\(=0+1\)
\(=1\)
=\(\left(\frac{15}{16}+\frac{1}{17^5}\right)+\left(\frac{16}{17}-\frac{15}{16}\right)\)
=\(\frac{1}{17^5}+\frac{16}{17}=\frac{1+16.17^4}{17^5}\)
đáp số là số thập phân vô hạn không tuần hoàn
\(\frac{121212}{171717}+\frac{2}{17}-\frac{40}{171}\)
\(=\frac{12}{17}+\frac{2}{17}-\frac{4}{17}=\frac{10}{17}\)
\(\Rightarrow A=B=\frac{10}{17}\)
121212/171717 + 2/7 - 40/171
= 12/17 + 2/7 - 4/17 = 10/17
= A = B = 10/17