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Zorluk: ZorEquivalent Algebraic Expressions

Which of the following expressions is equivalent to 2x23x9x29xx+3\frac{2x^2 - 3x - 9}{x^2 - 9} - \frac{x}{x+3} for all x>3x > 3?

  1. 11Cevap
  2. B
    x2+6x9x29\frac{x^2 + 6x - 9}{x^2 - 9}
  3. C
    x26x9x29\frac{x^2 - 6x - 9}{x^2 - 9}
  4. D
    3x+3x+3\frac{3x+3}{x+3}

Cevap

11
The correct answer is 11. Factoring the first term gives (2x+3)(x3)(x+3)(x3)\frac{(2x+3)(x-3)}{(x+3)(x-3)}. Since x>3x > 3, we cancel the common non-zero factor of x3x-3 to obtain 2x+3x+3\frac{2x+3}{x+3}. Subtracting xx+3\frac{x}{x+3} from this result yields 2x+3xx+3=1\frac{2x+3-x}{x+3} = 1.

Adım Adım Çözüm

1
Factor the numerator and the denominator of the first rational expression.
The numerator 2x23x92x^2 - 3x - 9 factors as (2x+3)(x3)(2x+3)(x-3) and the denominator x29x^2 - 9 factors as (x+3)(x3)(x+3)(x-3).
Factoring allows us to identify and divide out common factors to simplify the expression.
2
Simplify the first rational expression by cancelling the common factor.
For all x>3x > 3, the factor x3x-3 is non-zero, so the expression (2x+3)(x3)(x+3)(x3)\frac{(2x+3)(x-3)}{(x+3)(x-3)} simplifies to 2x+3x+3\frac{2x+3}{x+3}.
This simplifies the subtraction by reducing the first fraction.
3
Subtract the second expression from the simplified first expression.
2x+3x+3xx+3=(2x+3)xx+3=x+3x+3=1\frac{2x+3}{x+3} - \frac{x}{x+3} = \frac{(2x+3) - x}{x+3} = \frac{x+3}{x+3} = 1.
Since the denominators are identical, the numerators can be combined directly.

Anahtar Kavram

Simplifying rational expressions by factoring and performing algebraic operations.

Alternatif Yöntem

Instead of simplifying the first fraction first, find a common denominator immediately by multiplying the numerator and denominator of the second fraction by x3x-3. This yields: 2x23x9x29x(x3)x29=2x23x9x2+3xx29=x29x29=1\frac{2x^2 - 3x - 9}{x^2 - 9} - \frac{x(x-3)}{x^2 - 9} = \frac{2x^2 - 3x - 9 - x^2 + 3x}{x^2 - 9} = \frac{x^2 - 9}{x^2 - 9} = 1.
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