Решите уравнение: 1) (14х2 -49х)х-(2х-7)8х=0 2) (125х-25х2) 9х - (15х - 3х2)х=0 3) (0,81у2-0,9у) 0,9у=(0,1-0,09у)10у 4) (3/4у2 - 9/16у) 8у=1/7у2-3/28у

Ответ:

dina08254

08.10.2020 21:56

1.
$(14 {x}^{2} - 49x) \times x - (2x - 7) \times 8x = 0 \\ 7 {x}^{2} (2x - 7) - (2x - 7) \times 8x = 0 \\ (2x - 7)(7 {x}^{2} - 8x) = 0 \\(2x - 7)(7x - 8)x = 0 \\ - - - - - \\ 2x - 7 = 0 \\ 2x = 7 \\ x1 = \frac{7}{2} = 3.5 \\ - - - - \\ 7x - 8 = 0 \\ 7x = 8 \\ x2 = \frac{8}{7} = 1 \frac{1}{7} \\ - - - \\ x3 = 0$

2.
$(125x - 25 {x}^{2} ) \times 9x - (15x - 3 {x}^{2} ) \times x = 0 \\ 225 {x}^{2} \times (5 - x) - 3 {x}^{2} (5 - x) = 0 \\ (5 - x)(225 {x}^{2} - 3 {x}^{2} ) = 0 \\ (5 - x) \times 222 {x}^{2} = 0 \\ - - - \\ 5 - x = 0 \\ x1 = 5 \\ - - - \\ 222 {x}^{2} = 0 \\ x2 = 0$

3.
$(0.81 {y}^{2} - 0.9y) \times 0.9y = (0.1 - 0.09y) \times 10y \\ 0.81 {y}^{2} (0.9y - 1) - y + 0.9 {y}^{2} = 0 \\ 0.81 {y}^{2} (0.9y - 1) + y(0.9y - 1)= 0 \\ (0.9y - 1)(0.81y + 1)y = 0 \\ - - - \\ y1 = 0 \\ - - - \\ 0.9y - 1 = 0 \\ 0.9y = 1 \\ y2 = 1 \frac{1}{9} \\ - - - \\ 0.81y + 1 = 0 \\ 0.81y = - 1 \\ y3 = - 1 \frac{19}{81}$
4.
$( \frac{3}{4} {y}^{2} - \frac{9}{16} y) \times 8y = \frac{1}{7} {y}^{2} - \frac{3}{28} y \\ (y - \frac{3}{4} ) \times 6 {y}^{2} = (y - \frac{3}{4} ) \times \frac{1}{7} y \\ (y - \frac{3}{4} ) \times 6 {y}^{2} - (y - \frac{3}{4} ) \times \frac{1}{7} y = 0 \\ (y - \frac{3}{4} )(6 {y} - \frac{1}{7} ) \times y = 0 \\ - - - - \\ y1 = 0 \\ - - - \\ y - \frac{3}{4} = 0 \\ y2 = \frac{3}{4} \\ - - - \\ 6y - \frac{1}{7} = 0 \\ 6y = \frac{1}{7} \\ y3 = \frac{1}{42}$
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