Question

Cho biểu thức: P = \(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) với x > 0 và x \(\ne\) 1

a) Rút gọn P

b) Tính giá trị của P khi x = \(\frac{2}{2+\sqrt{3}}\)

c) Tìm x thỏa mãn \(P\sqrt{x}=6\sqrt{x}-3\)

Answer 1

a) P=\((\frac{x-1}{\sqrt{x}})\):\([\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}]\) (x>0;x≠1)

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

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

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

P=\(\sqrt{x}+1\)

Answer 2

b,Có x=\(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)

=>\(\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)

Có P=\(\sqrt{x}+1=\sqrt{3}-1+1=\sqrt{3}\)

c, Có P\(\sqrt{x}=6\sqrt{x}-3\)

<=>\(\sqrt{x}\left(P-6\right)+3=0\) <=> \(\sqrt{x}\left(\sqrt{x}+1-6\right)+3=0\) <=> \(\sqrt{x}\left(\sqrt{x}-5\right)+3=0\)

<=> \(x-5\sqrt{x}+3=0\) <=> \(\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) => \(\left[{}\begin{matrix}x=\frac{5+\sqrt{13}}{2}\left(tm\right)\\x=\frac{5-\sqrt{13}}{2}\left(tm\right)\end{matrix}\right.\)