Edited by vantho302, 04-05-2012 - 06:16.
\[\left( {\frac{{{a^3}}}{{{b^2}}} - a + \frac{1}{{{a^2}}}} \right){x^2} + b\left( {{b^2} - 1} \right)\]
Started By vantho302, 04-05-2012 - 06:15
#1
Posted 04-05-2012 - 06:15
vantho302
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Cho \[a > b > 0\]. Chứng minh rằng phương trình \[\left( {\frac{{{a^3}}}{{{b^2}}} - a + \frac{1}{{{a^2}}}} \right){x^2} + b\left( {{b^2} - 1} \right)x - 1 = 0\] có nghiệm thuộc \[\left( {0;1} \right)\]
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