Question

решите уравнение:

(2x^2-16x+11)/(x-7) + (5x-36)/(x-8)=2x+3

Answer 1

Замечательно!

Умножим все члены на (x-7)(x-8), получаем

2x³-16x²+11x-16x²+128x-88+5x²-36x-35x+252=2x³-30x²+112x+3x²-45x+168

После приведения подобных

x=4

Answer 2

$frac{2 {x}^{2} - 16x + 11 }{x - 7} + frac{5x - 36}{x - 8} = 2x + 3$
х ≠ 7, х ≠ 8
$frac{2 {x}^{2} - 16x + 11}{x - 7} + frac{5x - 36}{x - 8} = 2x + 3 \ frac{2 {x}^{2} - 16 + 11}{x - 7} + frac{5x - 36}{x - 8} - 2x = 3 \ frac{(2 {x}^{2} - 16x + 11)(x - 8) + (5x - 36)(x - 7) - 2(x - 7)(x - 8)}{(x - 7)(x - 8)} = 3 \ frac{2 {x}^{3} - 16 {x}^{2} + 11x - 16 {x}^{2} + 128x - 88 + 5 {x}^{2} - 36x - 35x + 252 - 2 {x}^{3} + 16 {x}^{2} + 14 {x}^{2} - 112x}{(x - 7)(x - 8)} = 3 \ frac{3 {x}^{2} - 44x + 164}{(x - 7)(x - 8)} = 3 \ 3 {x}^{2} - 44x + 164 = 3(x - 7)(x - 8) \ 3 {x}^{2} - 44x + 164 = (3x - 21)(x - 8) \ 3 {x}^{2} - 44x + 164 = 3 {x}^{2} - 24x - 21x + 168 \ 3 {x}^{2} - 44x - 3 {x}^{2} + 24x + 21x = 168 - 164$
х = 4, х ≠ 7, х ≠ 8
Ответ: х = 4