Đáp án:
a) \(A = \dfrac{{15}}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 121\\
\to \sqrt x = 11\\
\to A = \dfrac{{3.11 - 3}}{{11 - 3}} = \dfrac{{15}}{4}\\
b)B = \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{3}{{\sqrt x + 3}}\\
P = A.B = \dfrac{{3\sqrt x - 3}}{{\sqrt x - 3}}.\dfrac{3}{{\sqrt x + 3}} = \dfrac{{9\sqrt x - 9}}{{x - 9}}\\
d)P = \dfrac{3}{4}\\
\to \dfrac{{9\sqrt x - 9}}{{x - 9}} = \dfrac{3}{4}\\
\to 36\sqrt x - 36 = 3x - 27\\
\to 3x - 36\sqrt x + 9 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 6 + \sqrt {33} \\
\sqrt x = 6 - \sqrt {33}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 69 + 12\sqrt {33} \\
x = 69 - 12\sqrt {33}
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
c)P < 0\\
\to \dfrac{{9\sqrt x - 9}}{{x - 9}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
9\sqrt x - 9 < 0\\
x - 9 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
9\sqrt x - 9 > 0\\
x - 9 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt x < 1\\
x > 9
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
\sqrt x > 1\\
x < 9
\end{array} \right.
\end{array} \right.\\
\to 1 < x < 9
\end{array}\)
