Question
Solve:
$\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}$
$\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}$
✓
Answer
$\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}$
Let $\sqrt{\frac{x}{x-3}}=y$
Then $y +\frac{1}{ y }=\frac{5}{2}$
$\Rightarrow \frac{ y ^2+1}{ y }=\frac{5}{2}$
$\Rightarrow 2 y 2+2=5 y $
$ \Rightarrow 2 y 2-5 y+2=0$
$ \Rightarrow 2 y 2-4 y-y+2=0 $
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1)=0$
If $y-2=0$ or $2 y-1=0$ Then $y =2$ or $y =\frac{1}{2}$
$\Rightarrow \sqrt{\frac{x}{x-3}}=2$ or $\sqrt{\frac{x}{x-3}}=\frac{1}{2}$
$\Rightarrow \frac{x}{x-3}=4$ or $\frac{x}{x-3}=\frac{1}{4}$
$\Rightarrow x=4$ or $x=-1$
Let $\sqrt{\frac{x}{x-3}}=y$
Then $y +\frac{1}{ y }=\frac{5}{2}$
$\Rightarrow \frac{ y ^2+1}{ y }=\frac{5}{2}$
$\Rightarrow 2 y 2+2=5 y $
$ \Rightarrow 2 y 2-5 y+2=0$
$ \Rightarrow 2 y 2-4 y-y+2=0 $
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1)=0$
If $y-2=0$ or $2 y-1=0$ Then $y =2$ or $y =\frac{1}{2}$
$\Rightarrow \sqrt{\frac{x}{x-3}}=2$ or $\sqrt{\frac{x}{x-3}}=\frac{1}{2}$
$\Rightarrow \frac{x}{x-3}=4$ or $\frac{x}{x-3}=\frac{1}{4}$
$\Rightarrow x=4$ or $x=-1$
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