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Xét dấu các biểu thức sau :
a. \(\dfrac{{7x – 4}}{{8x + 5}} – 2\)
b. \(\dfrac{{{x^2} – 5x + 4}}{{{x^2} + 5x + 4}}\)
c. \(\dfrac{{15{x^2} – 7x – 2}}{{6{x^2} – x + 5}}\)
d. \(\dfrac{{{x^4} – 17{x^2} + 60}}{{x\left( {{x^2} – 8x + 5} \right)}}\)
:
a. Nếu đặt \(f\left( x \right) = \dfrac{{7x – 4}}{{8x + 5}} – 2\) thì
\(\begin{array}{l}f\left( x \right) > 0 \Leftrightarrow x \in \left( { – \dfrac{{14}}{9}; – \dfrac{5}{8}} \right)\\f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{{14}}{9}} \right) \cup \left( { – \dfrac{5}{8}; + \infty } \right).\end{array}\)
b. Nếu đặt \(g\left( x \right) = \dfrac{{{x^2} – 5x + 4}}{{{x^2} + 5x + 4}}\) thì
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\(\begin{array}{l}g\left( x \right) < 0 \Leftrightarrow x \in \left( { – 4; – 1} \right) \cup \left( {1;4} \right)\\g\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – 4} \right) \cup \left( { – 1;1} \right) \cup \left( {4; + \infty } \right).\end{array}\)
c. Nếu đặt \(h\left( x \right) = \dfrac{{15{x^2} – 7x – 2}}{{6{x^2} – x + 5}}\) thì
\(\begin{array}{l}h\left( x \right) > 0 \Leftrightarrow x \in \left( { – \infty ; – \dfrac{1}{5}} \right) \cup \left( {\dfrac{2}{3}; + \infty } \right)\\h\left( x \right) < 0 \Leftrightarrow x \in \left( { – \dfrac{1}{5};\dfrac{2}{3}} \right).\end{array}\)
d. Nếu đặt \(p\left( x \right) = \dfrac{{{x^4} – 17{x^2} + 60}}{{x\left( {{x^2} – 8x + 5} \right)}}\) thì \(p(x) > 0\) khi và chỉ khi
\(x \in \left( { – \sqrt {12} , – \sqrt 5 } \right) \cup \left( {0;4 – \sqrt {11} } \right)\)\( \cup \left( {\sqrt 5 ;\sqrt {12} } \right)\)\( \cup \left( {4 + \sqrt {11} ; + \infty } \right).\)
\(p(x) < 0\) khi và chỉ khi
\(x \in \left( { – \infty ; – \sqrt {12} } \right) \cup \left( { – \sqrt 5 ;0} \right)\)\( \cup \left( {4 – \sqrt {11} ;\sqrt 5 } \right) \cup \left( {\sqrt {12} ;4 + \sqrt {11} } \right).\)