Question

Решите уравнение: 1)\frac{3x-7}{x-1}- \frac{x+1}{x-1} =0 2)\frac{x}{x+5}- \frac{25}{x^{2}+5x } [/tex]

Answer 1

$1) \frac{3x - 7}{x - 1} - \frac{x + 1}{x - 1} = 0 \\ 3x - 7 - x - 1 = 0 \\ 2x = 8 \\ x = 4$
$2) \frac{x}{x + 5} - \frac{25}{ {x}^{2} + 5x} = 0 \\ {x}^{2} - 25 = 0 \\ x = 5$

Answer 2

1. Let's solve the equation:

$\frac{3x-7}{x-1} - \frac{x+1}{x-1} = 0$

To begin, let's simplify the expression by finding a common denominator:

$\frac{(3x-7) - (x+1)}{x-1} = 0$

Simplifying further:

$\frac{3x-7-x-1}{x-1} = 0$

Combining like terms:

$\frac{2x-8}{x-1} = 0$

Next, we can factor out the common factor of 2:

$\frac{2(x-4)}{x-1} = 0$

Now, we have two possibilities for the equation to hold true:

1. $2 = 0$ (which is not true)
2. $x - 4 = 0$

Solving the second equation:

$x - 4 = 0$ $x = 4$

Therefore, the solution to the equation is $x = 4$.

2. Let's solve the equation:

$\frac{x}{x+5} - \frac{25}{x^2+5x} = 0$

First, let's simplify the expression by finding a common denominator:

$\frac{x(x^2+5x)-25(x+5)}{(x+5)(x^2+5x)} = 0$

Expanding and combining like terms:

$\frac{x^3 + 5x^2 - 25x - 125}{x^3 + 10x^2 + 25x} = 0$

Next, we can factor the numerator:

$\frac{(x - 5)(x^2 + 10x + 25)}{x(x^2 + 10x + 25)} = 0$

Now, we have two possibilities for the equation to hold true:

1. $x - 5 = 0$ $x = 5$

2. $x^2 + 10x + 25 = 0$

This is a quadratic equation, and we can solve it by factoring:

$(x + 5)(x + 5) = 0$ $(x + 5)^2 = 0$ $x + 5 = 0$ $x = -5$

Therefore, the solutions to the equation are $x = 5$ and $x = -5$.