Question

3/х-6/x^2-3x=3x-7/3-x​

Answer 1

как я поняла формула

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у нас получается это

$\frac{9x}{3x^{2} } -\frac{18}{3x^{2} } -\frac{9x^{3} }{3x^{2} } =\frac{9x^{3} }{3x^{2} } -\frac{7x^{2} }{3x^{2} } -\frac{3x^{3} }{3x^{2} }$

теперь откидываем знаменатель

$9x - 18-9x^{3} =9x^{3}-7x^{2}-3x^{3}$

$9x -9x^{3}-9x^{3}+7x^{2}+3x^{3} =18$

известные в право, неизвестные в лево

приводим подобные

$9x -15x^{3}+7x^{2} =18$      

а дальше я не знаю

Answer 2

To solve the given equation, let's first simplify it by finding a common denominator for both sides. The common denominator will be x^2 - 3x.

The equation is: (3/x) - (6/x^2) - (3x) = (3x - 7)/(3 - x)

Multiplying each term by the common denominator, we get: 3(x^2 - 3x)/x(x^2 - 3x) - 6(x^2 - 3x)/x^2(x^2 - 3x) - 3x(x^2 - 3x)/x(x^2 - 3x) = (3x - 7)/(3 - x)

Simplifying, we have: 3(x^2 - 3x) - 6(x^2 - 3x) - 3x(x^2 - 3x) = (3x - 7)(x^2 - 3x)

Expanding the expressions: 3x^2 - 9x - 6x^2 + 18x - 3x^3 + 9x^2 = 3x^3 - 9x^2 - 7x^2 + 21x - 3x^2 + 9x

Combining like terms: -3x^3 - 9x^2 + 9x^2 + 7x^2 - 21x + 3x^2 - 9x + 9x = 0

Simplifying further: -3x^3 - 7x^2 - 21x = 0

Now, let's solve for x. Factoring out an x from the equation: x(-3x^2 - 7x - 21) = 0

Either x = 0 or -3x^2 - 7x - 21 = 0

To solve the quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a)

For the quadratic equation -3x^2 - 7x - 21 = 0: a = -3, b = -7, c = -21

x = (-(-7) ± √((-7)^2 - 4(-3)(-21)))/(2(-3)) x = (7 ± √(49 - 252))/(-6) x = (7 ± √(-203))/(-6)

Since the discriminant is negative, √(-203) is imaginary. Therefore, there are no real solutions for the quadratic equation.

Therefore, the solutions to the given equation are x = 0 and there are no real solutions for the quadratic equation -3x^2 - 7x - 21 = 0.