Soru

Zorluk: OrtaProperties of Exponents in Algebraic Expressions

For all non-zero real numbers xx and yy, the expression (x2+y−1)2x4\frac{(x^2 + y^{-1})^2}{x^4} is equivalent to which of the following?

  1. A
    1+x−4y−21 + x^{-4}y^{-2}
  2. B
    1+2x−2y+x−4y21 + 2x^{-2}y + x^{-4}y^2
  3. 1+2x−2y−1+x−4y−21 + 2x^{-2}y^{-1} + x^{-4}y^{-2}Cevap
  4. D
    1+2x2y−1+y−21 + 2x^2 y^{-1} + y^{-2}
  5. E
    y−2+2x−2y−2+x−4y−2y^{-2} + 2x^{-2}y^{-2} + x^{-4}y^{-2}

Cevap

1+2x−2y−1+x−4y−21 + 2x^{-2}y^{-1} + x^{-4}y^{-2}
The correct expression is obtained by expanding the binomial in the numerator using the perfect square formula (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2. Substituting a=x2a = x^2 and b=y−1b = y^{-1} yields (x2+y−1)2=(x2)2+2(x2)(y−1)+(y−1)2=x4+2x2y−1+y−2(x^2 + y^{-1})^2 = (x^2)^2 + 2(x^2)(y^{-1}) + (y^{-1})^2 = x^4 + 2x^2 y^{-1} + y^{-2}. Dividing each term by the denominator x4x^4 and applying the quotient rule for exponents xmxn=xm−n\frac{x^m}{x^n} = x^{m-n} results in 1+2x−2y−1+x−4y−21 + 2x^{-2}y^{-1} + x^{-4}y^{-2}.

Adım Adım Çözüm

1
Expand the binomial in the numerator.
(x2+y−1)2=(x2)2+2(x2)(y−1)+(y−1)2=x4+2x2y−1+y−2(x^2 + y^{-1})^2 = (x^2)^2 + 2(x^2)(y^{-1}) + (y^{-1})^2 = x^4 + 2x^2 y^{-1} + y^{-2}
Apply the binomial expansion formula (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 where a=x2a = x^2 and b=y−1b = y^{-1}.
2
Divide each term in the expanded numerator by the denominator.
x4x4+2x2y−1x4+y−2x4\frac{x^4}{x^4} + \frac{2x^2 y^{-1}}{x^4} + \frac{y^{-2}}{x^4}
Distribute the division over the terms in the numerator.
3
Simplify each fraction using exponent properties.
1+2x−2y−1+x−4y−21 + 2x^{-2}y^{-1} + x^{-4}y^{-2}
Use the quotient rule xmxn=xm−n\frac{x^m}{x^n} = x^{m-n} to simplify the variables.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Bu soruyu puanla