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Didi12342
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Fie nr rational
a = 1/2 × 3/4 × 5/6 ×...× 9999/10000
Aratati ca a < 0,01

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Răspuns :

Rayzen
[tex] a = \dfrac{1}{2} \times \dfrac{3}{4}\times \dfrac{5}{6} \times ... \times \dfrac{9999}{10000} \\ \\ $Fie: $\\ \\ b = \dfrac{2}{3} \times \dfrac{4}{5}\times \dfrac{6}{7} \times ... \times \dfrac{9998}{9999} \times \dfrac{10000}{10001} \\ \\ \\ \left(\dfrac{1}{2}< \dfrac{2}{3}\right), \left(\dfrac{3}{4} < \dfrac{4}{5}\right) ,..., \left(\dfrac{9999}{10000}<\dfrac{10000}{10001}\right) \\ \\ $Demonstratie:$ \\ \\ \dfrac{k}{k+1} < \dfrac{k+1}{k+2}, $ $ (k > 0)\Rightarrow \\ \\ \Rightarrow k(k+2) < (k+1)(k+1) \Rightarrow \\ \\ \Rightarrow k^2 + 2k < k^2+2k+1 \Rightarrow 0< 1 \quad (A) \\ \\ \\ ~~~~~~\dfrac{1}{2}< \dfrac{2}{3} \\ \\ ~~~~~~\dfrac{3}{4} < \dfrac{4}{5} \\ ~~~~~~~~~~\vdots \\ \\ \dfrac{9999}{10000}<\dfrac{10000}{10001} \\ --------$ $(\times) \\ \\ \Rightarrow \dfrac{1}{2} \times \dfrac{3}{4}\times \dfrac{5}{6} \times ... \times \dfrac{9999}{10000} < \\ \\ <\dfrac{2}{3} \times \dfrac{4}{5}\times \dfrac{6}{7} \times ... \times \dfrac{9998}{9999} \times \dfrac{10000}{10001}[/tex]

[tex]\Rightarrow a < b\Big| \times a,$ $ (a > 0) \Rightarrow a\times a < a \times b\\ \\ a \times b = \Big(\dfrac{1}{2} \times \dfrac{3}{4}\times \dfrac{5}{6} \times ... \times \dfrac{9999}{10000}\Big) \times \\ \\ \times \Big(\dfrac{2}{3} \times \dfrac{4}{5}\times \dfrac{6}{7} \times ... \times \dfrac{9998}{9999} \times \dfrac{10000}{10001}\Big) = \\ \\ = \dfrac{1}{10001}\\ \\ \Rightatrow a\times b < \dfrac{1}{10001} < \dfrac{1}{10000} \\ \\ a\times a< a\times b \Rightarrow a\times a < \dfrac{1}{10000} \Rightarrow \\ \\ \Rightarrow a\times a < \dfrac{1}{100} \times \dfrac{1}{100} \Rightarrow a < \dfrac{1}{100} \Rightarrow \\ \\ \Rightarrow \boxed{a < 0,01}[/tex]
observând că numărătorii sunt impari
a=1/2×3/5×.........×9999/10000
fie b format din numărătorii pari
b=2/3×4/5×...........×10000/10001
a×b=1/2×2/3×3/4×............×9999/10000×10000/10001
1/2×2/3(se simplifica 2)urmând să se simplifice toți ceilalți termeni ai fractiilor rămânând a×b=1/10001
a<b|×a
a×a<a×b
a×a<1/10001
cum 1/10001<1/10000
a×a<1/10000
a×a<0,01×0,01
a<0,01

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