Question

1/x+4-1/x+6=1/4 5х+8/х+6=3х+4/х+3

Answer 1

Ответ:

Пошаговое объяснение:

$1)\frac{1}{x}+4-\frac{1}{x}+6=\frac{1}{4}\\\\10\neq\frac{1}{4}\\\\2)5x+\frac{8}{x}+6=3x+\frac{4}{x}+3\\\\5x+\frac{8}{x}-3x-\frac{4}{x}=3-6\\\\2x+\frac{4}{x}=-3\\\\2x^{2}+3x+4=0$

D = b² - 4ac = 9 - 4*2*4 = 9 - 32 = -23

D < 0 корней нет

Answer 2

To solve the equations:

1. 1/(x+4) - 1/(x+6) = 1/4

Let's find a common denominator for the fractions on the left-hand side of the equation, which is (x+4)(x+6). Multiplying each term by this denominator gives us:

[(x+6) - (x+4)]/(x+4)(x+6) = 1/4

Simplifying the numerator:

[x + 6 - x - 4]/(x+4)(x+6) = 1/4

2/(x+4)(x+6) = 1/4

Cross-multiplying:

4 * 2 = (x+4)(x+6)

8 = (x+4)(x+6)

Expanding the right-hand side:

8 = x^2 + 10x + 24

Rearranging the equation:

x^2 + 10x + 24 - 8 = 0

x^2 + 10x + 16 = 0

To factor this quadratic equation, we look for two numbers that multiply to give 16 and add up to 10. The numbers 2 and 8 fit this criteria:

(x + 2)(x + 8) = 0

Setting each factor equal to zero:

x + 2 = 0 or x + 8 = 0

Solving for x:

x = -2 or x = -8

Therefore, the solutions for the first equation are x = -2 and x = -8.

2. (5x + 8)/(x + 6) = (3x + 4)/(x + 3)

Cross-multiplying:

(x + 6)(3x + 4) = (x + 3)(5x + 8)

Expanding both sides:

3x^2 + 4x + 18x + 24 = 5x^2 + 8x + 15x + 24

Combining like terms:

3x^2 + 22x + 24 = 5x^2 + 23x + 24

Subtracting (5x^2 + 23x + 24) from both sides:

0 = 5x^2 + 23x + 24 - 3x^2 - 22x - 24

0 = 2x^2 + x

Factoring out x:

0 = x(2x + 1)

Setting each factor equal to zero:

x = 0 or 2x + 1 = 0

If 2x + 1 = 0, then:

2x = -1 x = -1/2

Therefore, the solutions for the second equation are x = 0 and x = -1/2.