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\(C=\frac{3.8.15....80.99}{4.9.16.81.100}\)
\(=\frac{1.3.2.4.3.5...8.10.9.11}{2.2.3.3.4.4...9.9.10.10}\)
\(=\frac{\left(1.2.3....9\right).\left(3.4.5...10.11\right)}{\left(2.3.4.5...10\right).\left(2.3.4...10\right)}\)
\(=\frac{11}{10}\)
trả lời
c=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{99}{100}\)
C=\(\frac{3.8.15....99}{4.9.16.100}\)
C=\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\)
C=\(\frac{\left(1.2.....9\right)}{2.3....10}.\left(\frac{3.4....11}{2.3...10}\right)\)
C=\(\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
=\(\frac{2x4}{3x3}x\frac{3x5}{4x4}x\frac{4x6}{5x5}x...x\frac{9x11}{10x10}\)
=\(\frac{\left(2x3x4x5x6x..x9\right)x\left(4x5x6x...x11\right)}{\left(3x4x5x6x7x8x9x10\right)x\left(3x4x5x...x10\right)}\)
=\(\frac{2x11}{10x3}=\frac{22}{30}=\frac{11}{15}\)
Lời giải:
$A=(1-\frac{1}{4})+(1-\frac{1}{9})+(1-\frac{1}{16})+....+(1-\frac{1}{10000})$
$=(1+1+...+1)-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})$
$=99-(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....+\frac{1}{10000})< 99$
\(x=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{99}{100}=\dfrac{1.3}{2^2}+\dfrac{2.4}{3^2}+\dfrac{3.5}{4^2}+...+\dfrac{9.11}{10^2}=\dfrac{1.2.3...9}{2.3.4...10}.\dfrac{3.4.5...11}{2.3.4...10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)