Cho biểu thức T= ( căn bậc hai a + 1/ căn bậc hai a-1 - căn bậc hai a-1 / căn bậc hai a+ 1
1) \[T = \left( {\frac{{{{\left( {\sqrt a + 1} \right)}^2} - {{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}} \right){\left( {\frac{{\sqrt a }}{4} - \frac{1}{{4\sqrt a }}} \right)^2}\] |
\[ = \left( {\frac{{{{\left( {\sqrt a + 1} \right)}^2} - {{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}} \right){\left( {\frac{{a - 1}}{{4\sqrt a }}} \right)^2}\] |
\[ = \frac{{4\sqrt a }}{{a - 1}}.\frac{{{{\left( {a - 1} \right)}^2}}}{{{{\left( {4\sqrt a } \right)}^2}}}\] |
\[ = \frac{{a - 1}}{{4\sqrt a }}.\] |
2) \[\frac{{a - 1}}{{4\sqrt a }} = - \,\sqrt a - 1 \Leftrightarrow 5a + 4\sqrt a - 1 = 0\] |
\[\sqrt a = - 1\] hoặc \[\sqrt a = \frac{1}{5}\]. Kết luận \(a = \frac{1}{{25}}.\) |