Question

(2x^2+x-1)/(x^2-1)=0

Answer 1

Ответ:2X^2 + X - 1 \ (X-1)*(X+1)

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2X^2 + X - 1 = 0

D = 1 -4*2*(-1) = 9 V D = 3

X1 = - 1 + 3 \ 4 = 2\4 = 0.5

X2 = - 1 - 3 \ 4 = - 1

Теперь данное уравнение записываем следующим образом:

(X + 1)*(X - 0.5) \ (X - 1)*(X+1) = X - 0.5 \ X - 1

Объяснение:

Answer 2

Объяснение:

$\frac{2x^2+x-1}{x^2-1} =0.\\\frac{2x^2+x-1}{(x+1)(x-1)} =0.$

ОДЗ: х+1≠0     х≠-1    х-1≠0     х≠1.

$\frac{2x^2+2x-x-1}{(x+1)(x-1)}=0\\\frac{2*(x+1)-(x+1)}{(x+1)(x-1)}=0\\ \frac{(x+1)(2x-1)}{(x+1)(x-1)}=0\\ \frac{2x-1}{x-1}=0\\ 2x-1=0\\ 2x=1\ |:2\\x=0,5.$

Ответ: x=0,5.

Answer 3

To solve the equation (2x^2 + x - 1) / (x^2 - 1) = 0, you can start by factoring both the numerator and the denominator:

(2x^2 + x - 1) / ((x - 1)(x + 1)) = 0

Now, you have a fraction on the left-hand side, and a fraction is equal to zero when its numerator is zero while the denominator is nonzero. So, set the numerator equal to zero:

2x^2 + x - 1 = 0

Now, you can solve this quadratic equation for x. You can use the quadratic formula, which is:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = 1, and c = -1. Plug these values into the formula:

x = (-1 ± √(1^2 - 4(2)(-1))) / (2(2))

x = (-1 ± √(1 + 8)) / 4

x = (-1 ± √9) / 4

x = (-1 ± 3) / 4

Now, you have two possible solutions:

1. x = (-1 + 3) / 4 = 2 / 4 = 1/2
2. x = (-1 - 3) / 4 = -4 / 4 = -1

So, the solutions to the equation (2x^2 + x - 1) / (x^2 - 1) = 0 are x = 1/2 and x = -1.