Question

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

Answer 1

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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