Cho P=(căn x/căn x -1 – 1/căn x +1):(1/căn x +1 + 2/x-1) vs x>0,x khác 1 a)rút gọn P b) tính P khi x=4- 2 căn 3 10/07/2021 Bởi Claire Cho P=(căn x/căn x -1 – 1/căn x +1):(1/căn x +1 + 2/x-1) vs x>0,x khác 1 a)rút gọn P b) tính P khi x=4- 2 căn 3
$\begin{array}{l} P = \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x + 1}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{x – 1}}} \right)\\ P = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) – \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}:\left( {\dfrac{{\sqrt x – 1 + 2}}{{x – 1}}} \right)\\ P = \dfrac{{x + \sqrt x – \sqrt x + 1}}{{x – 1}}:\dfrac{{\sqrt x + 1}}{{x – 1}}\\ P = \dfrac{{x + 1}}{{\sqrt x + 1}}\\ b)\\ x = 4 – 2\sqrt 3 = {\left( {\sqrt 3 – 1} \right)^2}\\ \Rightarrow P = \dfrac{{4 – 2\sqrt 3 + 1}}{{\sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} + 1}} = \dfrac{{5 – 2\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{5\sqrt 3 – 6}}{3} \end{array}$ Bình luận
\(a) \\ P = ( \dfrac{\sqrt{x}}{\sqrt{x}-1} – \dfrac{1}{\sqrt{x}+1} ) : ( \dfrac{1}{\sqrt{x}+1} + \dfrac{2}{x-1} ) \\ = \dfrac{\sqrt{x}(\sqrt{x}+1)-(\sqrt{x}-1)}{x-1} : \dfrac{\sqrt{x}-1+2}{x-1} \\ = \dfrac{x+\sqrt{x}-\sqrt{x}+1}{x-1} : \dfrac{\sqrt{x}+1}{x-1} \\ = \dfrac{x+1}{x-1} : \dfrac{\sqrt{x}+1}{x-1} = \dfrac{x+1}{x-1} . \dfrac{x-1}{\sqrt{x}+1}\\=\dfrac{x+1}{\sqrt{x}+1}\\ b)\\ \text{Tại} \ x=4-2\sqrt{3}=(\sqrt{3}-1)^2 \ \text{giá trị của biểu thức P là :} \\ \dfrac{4-2\sqrt{3}+1}{\sqrt{(\sqrt{3}-1)^2}+1}=\dfrac{5-2\sqrt{3}}{|\sqrt{3}-1|+1}\\=\dfrac{5-2\sqrt{3}}{-1+\sqrt{3}+1}=\dfrac{5-2\sqrt{3}}{\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\) Bình luận
$\begin{array}{l} P = \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x + 1}}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{x – 1}}} \right)\\ P = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) – \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}:\left( {\dfrac{{\sqrt x – 1 + 2}}{{x – 1}}} \right)\\ P = \dfrac{{x + \sqrt x – \sqrt x + 1}}{{x – 1}}:\dfrac{{\sqrt x + 1}}{{x – 1}}\\ P = \dfrac{{x + 1}}{{\sqrt x + 1}}\\ b)\\ x = 4 – 2\sqrt 3 = {\left( {\sqrt 3 – 1} \right)^2}\\ \Rightarrow P = \dfrac{{4 – 2\sqrt 3 + 1}}{{\sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} + 1}} = \dfrac{{5 – 2\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{5\sqrt 3 – 6}}{3} \end{array}$
\(a) \\ P = ( \dfrac{\sqrt{x}}{\sqrt{x}-1} – \dfrac{1}{\sqrt{x}+1} ) : ( \dfrac{1}{\sqrt{x}+1} + \dfrac{2}{x-1} ) \\ = \dfrac{\sqrt{x}(\sqrt{x}+1)-(\sqrt{x}-1)}{x-1} : \dfrac{\sqrt{x}-1+2}{x-1} \\ = \dfrac{x+\sqrt{x}-\sqrt{x}+1}{x-1} : \dfrac{\sqrt{x}+1}{x-1} \\ = \dfrac{x+1}{x-1} : \dfrac{\sqrt{x}+1}{x-1} = \dfrac{x+1}{x-1} . \dfrac{x-1}{\sqrt{x}+1}\\=\dfrac{x+1}{\sqrt{x}+1}\\ b)\\ \text{Tại} \ x=4-2\sqrt{3}=(\sqrt{3}-1)^2 \ \text{giá trị của biểu thức P là :} \\ \dfrac{4-2\sqrt{3}+1}{\sqrt{(\sqrt{3}-1)^2}+1}=\dfrac{5-2\sqrt{3}}{|\sqrt{3}-1|+1}\\=\dfrac{5-2\sqrt{3}}{-1+\sqrt{3}+1}=\dfrac{5-2\sqrt{3}}{\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)