Đáp án:
a. \(\dfrac{{x + 8}}{{\sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
P = \dfrac{{x\sqrt x - 3 - \left( {2\sqrt x - 6} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - 3 - 2x + 12\sqrt x - 18 - x - 4\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - 3x + 8\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 8}}{{\sqrt x + 1}}\\
b.Thay:x = 14 - 6\sqrt 5 = 9 - 2.3.\sqrt 5 + 5\\
= {\left( {3 - \sqrt 5 } \right)^2}\\
\to P = \dfrac{{14 - 6\sqrt 5 + 8}}{{\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} + 1}} = \dfrac{{22 - 6\sqrt 5 }}{{3 - \sqrt 5 + 1}}\\
= \dfrac{{22 - 6\sqrt 5 }}{{4 - \sqrt 5 }}\\
c.P = - \dfrac{{17}}{4}\\
\to \dfrac{{x + 8}}{{\sqrt x + 1}} = - \dfrac{{17}}{4}\\
\to 4x + 32 = - 17\sqrt x - 17\\
\to 4x + 17\sqrt x + 49 = 0\left( {vô lý} \right)\\
Do:4x + 17\sqrt x + 49 > 0\forall x \ge 0;x \ne 9
\end{array}\)
⇒ Không tồn tại giá trị x TMĐK
