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

Answer 3

To solve the given equation, we need to find the value(s) of 'x' that satisfy the equation. To do that, let's first simplify the equation by getting rid of the denominators. The common denominator for (x-7) and (x-8) is (x-7)(x-8). So, let's multiply both sides of the equation by (x-7)(x-8) to eliminate the fractions.

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

Multiply both sides by (x - 7)(x - 8):

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

Now, simplify:

Cancel out the common factors:

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

Combine like terms:

2x^2 - 11x - 25 = (x - 7)(x - 8)(2x + 3)

Now, we have a quadratic equation. Let's set it to zero and solve for 'x':

2x^2 - 11x - 25 = 0

To solve this quadratic equation, we can use factoring or the quadratic formula. In this case, the quadratic equation does not factor easily, so we'll use the quadratic formula:

For an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 2, b = -11, and c = -25. Plugging these values into the formula:

x = (11 ± √((-11)^2 - 4 * 2 * (-25))) / 2 * 2 x = (11 ± √(121 + 200)) / 4 x = (11 ± √321) / 4

So, the solutions for 'x' are:

x = (11 + √321) / 4 x = (11 - √321) / 4

These are the two values of 'x' that satisfy the given equation.