Question

решите неравенство
ответ без логарифма должен быть

Answer 1

$\displaystyle\bf\\9^{x+1} -2\cdot 3^{x} < 7\\\\(3^{2}) ^{x+1} -2\cdot 3^{x} -7 < 0\\\\3^{2x}\cdot 3^{2} -2\cdot 3^{x} -7 < 0\\\\9\cdot 3^{2x} -2\cdot 3^{x} -7 < 0\\\\3^{x} =m \ \ ; \ \ m > 0\\\\9m^{2} -2m-7 > 0\\\\9m^{2} -2m-7=0\\\\D=(-2)^{2} -4\cdot 9\cdot(-7)=4+252=256=16^{2} \\\\\\m_{1} =\frac{2-16}{18} =-\frac{14}{18} =-\frac{7}{9} \\\\\\m_{2} =\frac{2+16}{18} =\frac{18}{18} =1\\\\\\\Big(m+\frac{7}{9}\Big)\cdot\Big(m-1\Big) < 0$

$\displaystyle\bf\\+ + + + + \Big(-\frac{7}{9} \Big)- - - - - \Big(1\Big)+ + + + + \\\\\\\left \{ {{m > -\dfrac{7}{9} } \atop {m < 1}} \right. \ \ \Rightarrow \ \ \left \{ {{3^{x} > -\dfrac{7}{9} } \atop {3^{x} < 1}} \right. \ \ \Rightarrow \ \ \left \{ {{x\in R } \atop {3^{x} < 3^\circ }} \right. \ \ \Rightarrow \ \ x < 0\\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ 0\Big)$