• $$\Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;}$$ $$\Large y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0$$
 A) If x>y B) $$\Large If \ x\ge y$$ C) lfx D) If x = y or relation cannot be established

 D) If x = y or relation cannot be established

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

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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