Представим последнее слагаемое в виде:
− 48 ⋅ 4 x 2 − 3 x = − 12 ⋅ 4 x 2 − 3 x + 1 -48\cdot4^{x^2 - 3x} = -12\cdot4^{x^2 - 3x + 1} − 48 ⋅ 4 x 2 − 3 x = − 12 ⋅ 4 x 2 − 3 x + 1 7 ⋅ 9 x 2 − 3 x + 1 + 5 ⋅ 6 x 2 − 3 x + 1 − 12 ⋅ 4 x 2 − 3 x + 1 = 0 7\cdot9^{x^2 -3x + 1} + 5\cdot6^{x^2 - 3x + 1} -12\cdot4^{x^2 - 3x + 1} = 0 7 ⋅ 9 x 2 − 3 x + 1 + 5 ⋅ 6 x 2 − 3 x + 1 − 12 ⋅ 4 x 2 − 3 x + 1 = 0 4 x 2 − 3 x + 1 4^{x^2 - 3x + 1} 4 x 2 − 3 x + 1 4 x 2 − 3 x + 1 > 0 4^{x^2 - 3x + 1} > 0 4 x 2 − 3 x + 1 > 0 7 ⋅ ( 9 4 ) x 2 − 3 x + 1 + 5 ⋅ ( 6 4 ) x 2 − 3 x + 1 − 12 = 0 7\cdot\left(\frac{9}{4}\right)^{x^2 -3x + 1} + 5\cdot\left(\frac{6}{4}\right)^{x^2 - 3x + 1} -12 = 0 7 ⋅ ( 4 9 ) x 2 − 3 x + 1 + 5 ⋅ ( 4 6 ) x 2 − 3 x + 1 − 12 = 0 7 ⋅ ( 3 2 ) 2 ( x 2 − 3 x + 1 ) + 5 ⋅ ( 3 2 ) x 2 − 3 x + 1 − 12 = 0 7\cdot\left(\frac{3}{2}\right)^{2(x^2 -3x + 1)} + 5\cdot\left(\frac{3}{2}\right)^{x^2 - 3x + 1} -12 = 0 7 ⋅ ( 2 3 ) 2 ( x 2 − 3 x + 1 ) + 5 ⋅ ( 2 3 ) x 2 − 3 x + 1 − 12 = 0 t = ( 3 2 ) x 2 − 3 x + 1 > 0 t = \left(\frac{3}{2}\right)^{x^2 - 3x + 1} > 0 t = ( 2 3 ) x 2 − 3 x + 1 > 0 7 t 2 + 5 t − 12 = 0 7t^2 + 5t - 12 = 0 7 t 2 + 5 t − 12 = 0 D = 25 + 336 = 361 D = 25 + 336 = 361 D = 25 + 336 = 361 t = − 5 ± 361 14 t = \frac{-5 \pm \sqrt{361}}{14} t = 14 − 5 ± 361 t = − 5 − 19 14 = − 24 14 = − 12 7 (не подходит, так как t > 0 ) t = \frac{-5 - 19}{14} = -\frac{24}{14} = -\frac{12}{7} \quad \text{(не подходит, так как \(t > 0\))} t = 14 − 5 − 19 = − 14 24 = − 7 12 ( не подходит , так как t > 0) t = − 5 + 19 14 = 14 14 = 1 t = \frac{-5 + 19}{14} = \frac{14}{14} = 1 t = 14 − 5 + 19 = 14 14 = 1
Обратная замена:
( 3 2 ) x 2 − 3 x + 1 = 1 ⇔ ( 3 2 ) x 2 − 3 x + 1 = ( 3 2 ) 0 \left(\frac{3}{2}\right)^{x^2 - 3x + 1} = 1 \Leftrightarrow \left(\frac{3}{2}\right)^{x^2 - 3x + 1} = \left(\frac{3}{2}\right)^0 ( 2 3 ) x 2 − 3 x + 1 = 1 ⇔ ( 2 3 ) x 2 − 3 x + 1 = ( 2 3 ) 0 x 2 − 3 x + 1 = 0 ⇔ x = 3 ± 5 2 x^2 - 3x + 1 = 0 \Leftrightarrow x = \frac{3 \pm \sqrt{5}}{2} x 2 − 3 x + 1 = 0 ⇔ x = 2 3 ± 5
б) Отберём корни, принадлежащие отрезку [ − 1 ; 2 ] [-1; 2] [ − 1 ; 2 ]
− 1 ≤ 3 − 5 2 ≤ 2 ⇔ − 5 ≤ − 5 ≤ 1 верно -1 \leq \frac{3 - \sqrt5}{2} \leq 2 \Leftrightarrow-5 \leq -\sqrt5 \leq 1 \text{ верно} − 1 ≤ 2 3 − 5 ≤ 2 ⇔ − 5 ≤ − 5 ≤ 1 верно − 1 ≤ 3 + 5 2 ≤ 2 ⇔ ≤ 2 ⇔ − 5 ≤ 5 ≤ 1 неверно -1 \leq \frac{3 + \sqrt5}{2} \leq 2 \Leftrightarrow \leq 2 \Leftrightarrow-5 \leq \sqrt5 \leq 1 \text{ неверно} − 1 ≤ 2 3 + 5 ≤ 2 ⇔≤ 2 ⇔ − 5 ≤ 5 ≤ 1 неверно
Ответ: а) 3 − 5 2 ; 3 + 5 2 ; \frac{3 - \sqrt{5}}{2}; \frac{3 + \sqrt{5}}{2}; 2 3 − 5 ; 2 3 + 5 ; 3 − 5 2 ; \frac{3 - \sqrt{5}}{2}; 2 3 − 5 ;
