tinh
\(\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{x}}}}}\)
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{100}\right)+x=2+\frac{1}{5}\)
\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}+x=\frac{11}{5}\)
\(\frac{1}{100}+x=\frac{11}{5}\)
\(x=\frac{11}{5}-\frac{1}{100}=\frac{219}{100}\)
ai làm được giúp mình nhé
A= (1-1/1x3)x(1-1/2x4)-(1-1/3x5)......x(1-1/20x22)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)
\(=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{x+1}{x}}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{x}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{x+1+x}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2x+1}{x+1}}}}=\frac{1}{1+\frac{1}{1+\frac{x+1}{2x+1}}}\)
\(=\frac{1}{1+\frac{1}{\frac{2x+1+x+1}{2x+1}}}=\frac{1}{1+\frac{1}{\frac{3x+2}{2x+1}}}=\frac{1}{1+\frac{2x+1}{3x+2}}=\frac{1}{\frac{3x+2+2x+1}{3x+2}}=\frac{1}{\frac{5x+3}{3x+2}}=\frac{3x+2}{5x+3}\)