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(x - 1)/(x - 3) - (x ^ 2 - x + 2)/(x ^ 2 - 2x - 3) = 1/(x + 1) prove it​

User MLKing
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2 Answers

5 votes

Answer:

Explanation:

(x-1)/(x-3)-(x²-x+2)/(x²-2x-3)=1/(x+1)


(x-1(x^(2)-2x-3)-x-3(x^(2)-x+2) )/(x-3(x^(2) -2x-3)) =0


((x^(3)-2x^(2) -3x-x^(2) +2x+3)-(x^(3)-x^(2) +2x-3x^(2) +3x-6) )/(x^(3)-2x^(2) -3x-3x^(2) +6x+9)=0


(-7x^(2) +2x+9)/(x^(3)-5x^(2) +3x+9 ) =0

-7x²+2x+9=x³-5x²+3x+9

x³-2x²-x=0

User Victor Oliveira
by
8.7k points
3 votes

Answer:

Bellow

Explanation:

To prove the equation:

(x - 1)/(x - 3) - (x ^ 2 - x + 2)/(x ^ 2 - 2x - 3) = 1/(x + 1)

we need to find a common denominator for the left-hand side of the equation. The common denominator is (x - 3)(x + 1)(x - 3), which is the product of all the factors in the denominators.

Using this common denominator, we can rewrite the left-hand side of the equation as:

[(x - 1)(x + 1)(x - 3) - (x ^ 2 - x + 2)(x + 1)] / [(x - 3)(x + 1)(x - 3)]

Simplifying the numerator using distributive property, we get:

[(x^3 - 2x^2 - 2x + 2) - (x^3 - 2x^2 - x - 2)] / [(x - 3)(x + 1)(x - 3)]

Simplifying the numerator further, we get:

(-x + 4) / [(x - 3)(x + 1)(x - 3)]

Now, we can simplify the right-hand side of the equation by multiplying both the numerator and the denominator by (x - 3)(x + 1), which gives us:

1 / (x + 1) = (x - 3)(x + 3) / [(x - 3)(x + 1)(x - 3)]

Therefore, the original equation can be written as:

(-x + 4) / [(x - 3)(x + 1)(x - 3)] = (x - 3)(x + 3) / [(x - 3)(x + 1)(x - 3)]

Cancelling the common factors on both sides, we get:

-x + 4 = x^2 - 9

Rearranging and simplifying, we get:

x^2 + x - 13 = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = (-1 ± √(1 + 4*13)) / 2

Simplifying, we get:

x = (-1 ± √53) / 2

Therefore, we have proved that the equation:

(x - 1)/(x - 3) - (x ^ 2 - x + 2)/(x ^ 2 - 2x - 3) = 1/(x + 1)

holds true for all values of x except (-1, 3, 3 - √13, 3 + √13).

User Nickolas George
by
9.1k points

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