Đáp án:
c) \(x = \dfrac{9}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{x\sqrt {2x} + 1 - x - \sqrt {2x} }}{{x - 1}}\\
= \dfrac{{\left( {x - 1} \right)\sqrt {2x} - \left( {x - 1} \right)}}{{x - 1}}\\
= \dfrac{{\left( {x - 1} \right)\left( {\sqrt {2x} - 1} \right)}}{{x - 1}}\\
= \sqrt {2x} - 1\\
B = \sqrt {4 + 2\sqrt 3 } - \dfrac{2}{{\sqrt 3 + 1}}\\
= \sqrt {3 + 2\sqrt 3 .1 + 1} - \dfrac{2}{{\sqrt 3 + 1}}\\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \dfrac{2}{{\sqrt 3 + 1}}\\
= \sqrt 3 + 1 - \dfrac{2}{{\sqrt 3 + 1}}\\
= \dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2} - 2}}{{\sqrt 3 + 1}} = \dfrac{{4 + 2\sqrt 3 - 2}}{{\sqrt 3 + 1}}\\
= \dfrac{{2 + 2\sqrt 3 }}{{\sqrt 3 + 1}} = 2\\
b)Thay:x = B = 2\\
\to A = \sqrt {2.2} - 1 = 1\\
c)A = B\\
\to \sqrt {2x} - 1 = 2\\
\to \sqrt {2x} = 3\\
\to 2x = 9\\
\to x = \dfrac{9}{2}
\end{array}\)
