Question

Cho biểu thức \(E=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\) \(\left(x>0;x\ne1\right)\)

a. Rút gọn E

b. Tìm x để E >0

Answer 1

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

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

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

= \(\frac{\sqrt{x}-1}{\sqrt{x}}\)

Answer 2

\(a,đkxđ:x>0,x\ne1\)

\(E=\frac{\sqrt{x}.\sqrt{x}}{(\sqrt{x}-1).\sqrt{x}}-\frac{2\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}\)

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

\(b,ĐểP>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}>0.do\sqrt{x}>0\Rightarrow\)\(\sqrt{x}-1>0\Rightarrow x>1\)

Answer 3

b) E>0

<=> \(\frac{\sqrt{x}-1}{\sqrt{x}}>0\)

<=> \(\sqrt{x}-1>0\)

<=> x>1