Question

I. $\frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x}$
II. $y^2-\frac{(11{)}^{\frac{5}{2}}}{\sqrt{y}}=0$

• A. $x>y$
• B. $x\ge y$
• C. $x<y$
• D. $x\le y$
• E. $x=y$or the relationship cannot be established

Answer 1

The correct option is E $x=y$or the relationship cannot be established

Equation I
$\frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x}\Rightarrow \frac{1}{\sqrt{x}}(4+7)=\sqrt{x}\Rightarrow \frac{11}{\sqrt{x}}=\sqrt{x}\Rightarrow 11=x$
Equation II
$y^2-\frac{(11{)}^{\frac{5}{2}}}{\sqrt{y}}=0$
$\Rightarrow (y{)}^2-\frac{(11{)}^{\frac{5}{2}}}{(y{)}^{\frac{1}{2}}}=0$
$\Rightarrow \frac{(y{)}^{\frac{5}{2}}-(11{)}^{\frac{5}{2}}}{(y{)}^{\frac{1}{2}}}=0\Rightarrow (y{)}^{\frac{5}{2}}=(11{)}^{\frac{5}{2}}\therefore y=11$
Hence, x = y