Đáp án:
\(\begin{array}{l}
a)x > 0;x \ne 1\\
b)E = 4x\\
c)8
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 1\\
b)E = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2} + 4\sqrt x \left( {x - 1} \right)}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x }}\\
= \dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1 + 4x\sqrt x - 4\sqrt x }}{{\sqrt x }}\\
= \dfrac{{4x\sqrt x }}{{\sqrt x }} = 4x\\
c)Thay:x = \left( {4 + \sqrt {15} } \right).\sqrt 2 .\left( {\sqrt 5 - \sqrt 3 } \right).\sqrt {4 - \sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {8 - 2\sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {5 - 2.\sqrt 5 .\sqrt 3 + 3} \\
= \left( {4 + \sqrt {15} } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left( {4 + \sqrt {15} } \right){\left( {\sqrt 5 - \sqrt 3 } \right)^2}\\
= \left( {4 + \sqrt {15} } \right)\left( {5 - 2\sqrt {15} + 3} \right)\\
= \left( {4 + \sqrt {15} } \right)\left( {8 - 2\sqrt {15} } \right)\\
= 2\left( {4 + \sqrt {15} } \right)\left( {4 - \sqrt {15} } \right)\\
= 2\left( {16 - 15} \right) = 2\\
\to E = 4.2 = 8
\end{array}\)
