Cho phương trình \(2\sqrt {8x} - 3\sqrt {2x} + 4\sqrt {32x} = 17\sqrt {x + 2} \). Khi đó:

a) Sai.

\(A\) xác định khi \(\frac{{10}}{{{x^3}}} \ge 0\), suy ra \(x > 0\). Vậy \(A\) xác định khi \(x > 0\).

b) Sai.

Với \(x > 0\) ta có: \(A = x\sqrt {\frac{{10}}{{{x^3}}}} = \sqrt {\frac{{10{x^2}}}{{{x^3}}}} = \sqrt {\frac{{10}}{x}} .\) Vậy \(A = \sqrt {\frac{{10}}{x}} .\)

c) Đúng.

\({\rm{B}} = 2\sqrt {1 + \frac{{\sqrt 3 }}{2}} = \sqrt {{2^2}\left( {1 + \frac{{\sqrt 3 }}{2}} \right)} = \sqrt {4 + 2\sqrt 3 } = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2 \cdot \sqrt 3 \cdot 1 + {1^2}} = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} = \sqrt 3 + 1.\)

Vậy \({\rm{B}} = \sqrt 3 + 1.\)

d) Sai.

Vì \({\rm{A}} = {\rm{B}} - \sqrt 3 \) nên \(\sqrt {\frac{{10}}{x}} = \sqrt 3 + 1 - \sqrt 3 \)

\(\sqrt {\frac{{10}}{x}} = 1\)

\(\frac{{10}}{x} = 1\)

\(x = 10\) (thỏa mãn).

Đáp án: 8

\(B = \frac{3}{{\sqrt 5 - \sqrt 2 }} + \frac{4}{{\sqrt 6 + \sqrt 2 }} + \frac{1}{{\sqrt 6 + \sqrt 5 }}\)

\( = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}} + \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{{\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 6 - \sqrt 2 } \right)}} + \frac{{\sqrt 6 - \sqrt 5 }}{{\left( {\sqrt 6 + \sqrt 5 } \right)\left( {\sqrt 6 - \sqrt 5 } \right)}}\)

\( = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{{{{\left( {\sqrt 6 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} + \frac{{\sqrt 6 - \sqrt 5 }}{{{{\left( {\sqrt 6 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}}\)

\( = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{3} + \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{4} + \frac{{\sqrt 6 - \sqrt 5 }}{1}\)

\( = \sqrt 5 + \sqrt 2 + \sqrt 6 - \sqrt 2 + \sqrt 6 - \sqrt 5 \)

\( = 2\sqrt 6 \).

Do đó, \(a = 2,\,\,b = 6\).

Vậy \(T = a + b = 8\).