Question

(5х-1)/( х+7)-(2х+2)/(х-3)+63/(х^2+4x-21)=0

Answer 1

X²+4x-21=0
x1+x2=-4 U x1*x2=-21⇒x1=-7 U x2=3
(5x-1)/(x+7)-(2x+2)/(x-3)=63/[9x+7)(x-3)]=0
x≠-7 x≠3
(5x-1)(x-3)-(2x+2)(x+7)-63=0
5x²-15x-x+3-2x²-14x-2x-14+63=0
3x²-32x+52=0
D=1024-624=400
x1=(32-20)/6=2
x2=(32+20)/6=26/3

Answer 2

To solve the equation $\frac{{5x - 1}}{{x + 7}} - \frac{{2x + 2}}{{x - 3}} + \frac{{63}}{{x^2 + 4x - 21}} = 0$, we need to find the values of $x$ that satisfy the equation. Let's begin by simplifying the equation.

First, we'll find a common denominator for the fractions: $\frac{{5x - 1}}{{x + 7}} - \frac{{2x + 2}}{{x - 3}} + \frac{{63}}{{x^2 + 4x - 21}} = 0$

The common denominator for the fractions is $(x + 7)(x - 3)(x^2 + 4x - 21)$.

Now, let's multiply each term by the common denominator to eliminate the fractions:

$(5x - 1)(x - 3)(x^2 + 4x - 21) - (2x + 2)(x + 7)(x^2 + 4x - 21) + 63(x + 7)(x - 3) = 0$

Expanding and simplifying:

$(5x - 1)(x^3 + 4x^2 - 21x - 3x^2 - 12x + 63) - (2x + 2)(x^3 + 4x^2 - 21x + 7x^2 + 28x - 147) + 63(x^2 + 4x - 21) = 0$

$(5x - 1)(x^3 + x^2 - 33x + 63) - (2x + 2)(x^3 + 11x^2 + 7x - 147) + 63x^2 + 252x - 1323 = 0$

Now, let's multiply out each term:

$5x^4 + 5x^3 - 165x^2 + 315x - x^3 - x^2 + 33x - 63 - 2x^4 - 22x^3 - 14x^2 + 294x + 63x^2 + 693x - 294 - 63x^2 - 252x + 1323 = 0$

Combining like terms:

$5x^4 - 2x^4 + 5x^3 - x^3 - 22x^3 - 165x^2 - x^2 - 14x^2 + 63x^2 + 63x^2 + 33x - 294x + 294x + 315x + 693x - 252x - 63 - 294 + 1323 = 0$

$3x^4 - 23x^3 - 113x^2 + 1095x + 966 = 0$

We now have a quartic equation. To solve it, we can use numerical methods such as factoring, graphing, or using a computer algebra system. However, solving this quartic equation algebraically can be quite complex and involve lengthy calculations. Therefore, I would recommend using numerical methods to approximate the solutions.