Đáp án:
a) \(\dfrac{{29\sqrt 6 }}{{18}}\)
b) x=2
Giải thích các bước giải:
\(\begin{array}{l}
a)3.\dfrac{{\sqrt 6 }}{2} - 2.\dfrac{{\sqrt 6 }}{6} + 4.\dfrac{{\sqrt 2 }}{{3\sqrt 3 }}\\
= \dfrac{{3\sqrt 6 }}{2} - \dfrac{{\sqrt 6 }}{3} + \dfrac{{4\sqrt 6 }}{9}\\
= \dfrac{{29\sqrt 6 }}{{18}}\\
b)DK:x \ge 2\\
\sqrt {x - 1} + \sqrt {x - 2} = \sqrt {2x - 3} \\
\to x - 1 + 2\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} + x - 2 = 2x - 3\\
\to 2x - 3 + 2\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} = 2x - 3\\
\to \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} = 0\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = 2
\end{array} \right.
\end{array}\)
