Question

Answer 1

Lời giải:

ĐKXĐ: $x>0$

\(P=\frac{x-1}{\sqrt{x}}: [\frac{(\sqrt{x}-1)(\sqrt{x}+1)+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}]\)

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

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

b. ĐKXĐ: $x\geq 4$

\(P\sqrt{x}=(\sqrt{x}+1)^2=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x+1+2\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x+4-4\sqrt{x}+\sqrt{x-4}=0\)

\(\Leftrightarrow (\sqrt{x}-2)^2+\sqrt{x-4}=0\)

\(\Rightarrow \sqrt{x}-2=\sqrt{x-4}=0\Leftrightarrow x=4\) (tm)

Vậy......