Đáp án:
\(\begin{array}{l}
1)23 - 9\sqrt 3 \\
2)a - \sqrt a \\
3)a = 4\\
4)Min = - \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:a > 0\\
1)A = \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}}\\
= \sqrt a \left( {\sqrt a + 1} \right) = a + \sqrt a \\
Thay:a = 19 - 8\sqrt 3 \\
= 16 - 2.4.\sqrt 3 + 3\\
= {\left( {4 - \sqrt 3 } \right)^2}\\
\to A = 19 - 8\sqrt 3 + \sqrt {{{\left( {4 - \sqrt 3 } \right)}^2}} \\
= 19 - 8\sqrt 3 + 4 - \sqrt 3 \\
= 23 - 9\sqrt 3 \\
2)B = \dfrac{{\sqrt a \left( {2\sqrt a + 1} \right)}}{{\sqrt a }} - 1\\
= 2\sqrt a + 1 - 1 = 2\sqrt a \\
A - B = a + \sqrt a - 2\sqrt a \\
= a - \sqrt a \\
3)A - B = 2\\
\to a - \sqrt a = 2\\
\to a - \sqrt a - 2 = 0\\
\to \left( {\sqrt a - 2} \right)\left( {\sqrt a + 1} \right) = 0\\
\to \sqrt a - 2 = 0\left( {do:\sqrt a + 1 > 0\forall a > 0} \right)\\
\to a = 4\\
4)A - B = a - \sqrt a \\
= a - 2\sqrt a .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {\sqrt a - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4}\\
Do:{\left( {\sqrt a - \dfrac{1}{2}} \right)^2} \ge 0\forall a > 0\\
\to {\left( {\sqrt a - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\to Min = - \dfrac{1}{4}\\
\Leftrightarrow \sqrt a - \dfrac{1}{2} = 0\\
\to a = \dfrac{1}{4}
\end{array}\)
