Rút gọn biểu thức \(B = \sqrt {x - \sqrt {{x^2} - 1} } .\sqrt {x + \sqrt {{x^2} - 1} } \) với x ≥ 1 ta được

Đáp án đúng là: A

Ta có: \(A = \frac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\)

\(A = \frac{{\sqrt {{x^3}} + \sqrt {{y^3}} }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\)

\(A = \frac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt x \sqrt y + y} \right)\)

\(A = x - \sqrt {xy} + y - x + 2\sqrt x \sqrt y - y\)

\(A = \sqrt {xy} \).

Đáp án đúng là: C

Ta có: \(A = \left( {4 + \sqrt {15} } \right)\left( {\sqrt {10} - \sqrt 6 } \right)\sqrt {4 - \sqrt {15} } \)

\(A = \left( {4\sqrt {10} - 4\sqrt 6 + \sqrt {15} .\sqrt {10} - \sqrt {15} .\sqrt 6 } \right)\sqrt {4 - \sqrt {15} } \)

\(A = \left( {4\sqrt {10} - 4\sqrt 6 + 5\sqrt 6 - 3\sqrt {10} } \right).\sqrt {4 - \sqrt {15} } \)

\(A = \left( {\sqrt {10} + \sqrt 6 } \right).\sqrt {4 - \sqrt {15} } \)

\(A = \sqrt 2 \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {4 - \sqrt {15} } \)

\(A = \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {2\left( {4 - \sqrt {15} } \right)} \)

\(A = \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {8 - 2\sqrt {15} } \)

\(A = \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \)

\(A = \left( {\sqrt 5 + \sqrt 3 } \right).\left( {\sqrt 5 - \sqrt 3 } \right) = {\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 3 } \right)^2} = 5 - 3 = 2\).