Cho hàm số $f(x) = x^2 + bx  +  1$. Với $b \in ( 3;\frac{7}{2})$ . Giải bất phương trình $f[ f(x)] > x$
Bất phương trình đã cho tương đương với $f(f(x)) - x > 0$
$ \Leftrightarrow {f^2} + bf + 1 - x + f - f > 0$
$ \Leftrightarrow {f^2} + bf + 1 + (f - x) - ({x^2} + b{\rm{x  +  1)}} > 0$
$ \Leftrightarrow {f^2} - {x^2} + b\left( {f - x} \right) + \left( {f - x} \right) > 0$
$ \Leftrightarrow \left( {f - x} \right)\left( {f + x + b + 1} \right) > 0$
$ \Leftrightarrow \left[ {{x^2} + \left( {b - 1} \right)x + 1} \right]\left[ {{x^2} + \left( {b + 1} \right)x + b + 2} \right] > 0$            (2)
Ký hiệu     ${f_1}(x) = {x^2} + \left( {b - 1} \right)x + 1$
${f_2}(x) = {x^2} + \left( {b + 1} \right)x + b + 2$
Tương ứng ta có     ${\Delta _1} = {\left( {b - 1} \right)^2} - 4 = {b^2} - 2b - 3$
${\Delta _2} = {\left( {b + 1} \right)^2} - 4\left( {b + 2} \right) = {\left( {b - 1} \right)^2} - 8$
Do $b \in \left( {3;7/2} \right)$ nên ${\Delta _2} < 0$ vậy (2) tương đương với ${f_1}(x) > 0$
${f_1}(x)$ có nghiệm $x = \frac{{1 - b - \sqrt {{\Delta _1}} }}{2},x = \frac{{1 - b + \sqrt {{\Delta _1}} }}{2}$
Từ đó nghiệm của (2) ( hay của (1) ) là  $x < \frac{{1 - b - \sqrt {{b^2} - 2b - 3} }}{2}$ hoặc $x < \frac{{1 - b + \sqrt {{b^2} - 2b - 3} }}{2}$

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