• \( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;} \) \( \Large y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0 \)

    A) If x>y

      B) \( \Large If \ x\ge y \)

    C) lfx

      D) If x = y or relation cannot be established

    Correct Answer:
      D) If x = y or relation cannot be established

    Description for Correct answer

    \( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x} = \frac{11}{\sqrt{x}} = \sqrt{x}\)

    Therefore, x = 11

    and \( \Large y^{2} - \frac{ \left(11\right)^{5/2} }{\sqrt{y}} = 0 = y^{2} = \frac{ \left(11\right)^{5/2} }{ \left(y\right)^{1/2} } \)

    = \( \Large y^{2} \times y^{1/2} = \left(11\right)^{5/2} \)

    = \( \Large \left(y\right)^{5/2} = \left(11\right)^{5/2} \)

    Therefore, y = 11

    Therefore, x = y


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