Решите уравнения используя метод введения новой переменной 1.9 (9-5x)^2+17 (9-5x)+8=02. 8...

1. 9(9-5x)² + 17(9-5x) + 8 = 0
(9-5x) - t
9t²+17t+8=0
D = b²-4ac = (17)²-4*9*8 = 289-288=1, √D = 1
$t_{1}= \frac{-b+ \sqrt{D} }{2a} = \frac{-17+1}{18}= \frac{-16}{18} = \frac{-8}{9}$
$t_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{-17-1}{18}= \frac{-18}{18} = -1$
Возвращаемся к замене:
9-5х = -8/9, -5х = -9 8/9, х = -9 8/9÷5 = - 1 44/45
9-5х = 1, -5х = 10, х = -2
Ответ: -1 44/45, -2

2. 8(10-3x)² - 5(10-3x)-3 = 0
(10-3x) - t
8t² -5t-3 = 0
D = b²-4ac = (-5)²-4*8*(-3) = 25+96 = 121,√D = 11
$t_{1} = \frac{-b- \sqrt{D} }{2a} = \frac{5-11}{16} = \frac{-6}{16}$
$t_{2} = \frac{-b+ \sqrt{D} }{2a} = \frac{5+11}{16} = \frac{16}{16} = 1$
Вовзращаемся к замене:
10-3x = -6/16, -3х = -10 3/8, х = 3 11/24
10-3х = 1, -3х = -9, х = 3
Ответ: 3, 3 11/24