Question

If $y\sqrt{x}=1+x$, then $\sqrt{x}{y}_1=$_____

• A. $\frac{x-1}{2x}$
• B. $\frac{1-x}{2x}$
• C. $\frac{x+1}{2x}$
• D. $\frac{x-1}{x}$

Answer 1

The correct option is A $\frac{x-1}{2x}$

$4\sqrt{x}=1+x$

$\sqrt{x}\cdot y1+y\cdot \frac{1}{2\sqrt{x}}=0+1$

$\sqrt{x}y1=1-\frac{y}{2\sqrt{x}}$

$=1-\frac{1}{2\sqrt{x}}\left(\frac{1+x}{\sqrt{x}}\right)$

$=1-\left(\frac{1+x}{2x}\right)$

$\sqrt{x}y1=\frac{x-1}{2x}$.