Question

P = ( \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2}{x-4}\)) : ( \(\sqrt{x}-1+\dfrac{\sqrt{x}-4}{\sqrt{x}}\)) Với x > 0 ; x khác 4

a) Chứng minh P = \(\sqrt{x}+3\)

b) Tìm x sao cho P = x + 3

Answer 1

\(a.P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2}{x-4}\right).\left(\sqrt{x}-1+\dfrac{\sqrt{x}-4}{\sqrt{x}}\right)=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{x-4}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-4}.\dfrac{x-4}{\sqrt{x}}=\sqrt{x}+3\left(x>0;x\ne4\right)\)

\(b.P=x+3\) ⇔ \(\sqrt{x}+3=x+3\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)

KL........