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Zorluk: OrtaSurds and Rationalization of Denominators

Simplify the surd expression 5+252\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}.

  1. 7+2103\frac{7 + 2\sqrt{10}}{3}Cevap
  2. B
    7+2103\frac{\sqrt{7} + 2\sqrt{10}}{3}
  3. C
    7+103\frac{7 + \sqrt{10}}{3}
  4. D
    7+2107\frac{7 + 2\sqrt{10}}{7}

Cevap

7+2103\frac{7 + 2\sqrt{10}}{3}
Multiplying both the numerator and denominator by the conjugate of the denominator, (5+2)(\sqrt{5} + \sqrt{2}), expands the numerator to 5+210+2=7+2105 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10} and simplifies the denominator using the difference of squares to 52=35 - 2 = 3, giving 7+2103\frac{7 + 2\sqrt{10}}{3}.

Adım Adım Çözüm

1
Identify the conjugate of the denominator
The conjugate of (52)(\sqrt{5} - \sqrt{2}) is (5+2)(\sqrt{5} + \sqrt{2}).
Rationalizing a binomial denominator requires multiplying by its conjugate to apply the difference of squares identity (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2.
2
Multiply both numerator and denominator by the conjugate
\frac{(\sqrt{5} + \sqrt{2})(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})}
This maintains the value of the fraction while removing radical terms from the denominator.
3
Expand the numerator and denominator
Numerator: (5)2+252+(2)2=5+210+2=7+210(\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}. Denominator: (5)2(2)2=52=3(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3.
Apply algebraic expansion (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 and difference of squares.
4
Combine terms to form the final simplified expression
7+2103\frac{7 + 2\sqrt{10}}{3}
The expression is now fully rationalized and in standard simplified surd form.

Anahtar Kavram

Rationalization of Binomial Denominators
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