Question

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

a,Rút gọn P với x>0,x khác 4 và x khác 1

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

Answer 1

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

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

=\(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

=\(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)(*)

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

\(\Rightarrow\sqrt{x}=\sqrt{2}-1\)

Thay vào (*) ta đc

\(\frac{\sqrt{2}-1-2}{3\sqrt{2}-3}=\frac{\sqrt{2}-3}{3\left(\sqrt{2}-1\right)}\)

Answer 2

\( a)P = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\ = \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{3}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\ = \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{3}\\ = \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\ b)x = 3 - 2\sqrt 2 \Rightarrow P = \dfrac{{\sqrt {3 - 2\sqrt 2 } - 2}}{{3\sqrt {3 - 2\sqrt 2 } }} = \dfrac{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} - 2}}{{3\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} }} = \dfrac{{\sqrt 2 - 3}}{{3\sqrt 2 - 3}} = \dfrac{{ - 1 - 2\sqrt 2 }}{3} \)