Cho:
$A=\frac{1}{\sqrt{x}+2}$ – $\frac{5}{x-\sqrt{x}-6}$ – $\frac{\sqrt{x}-2}{3-\sqrt{x}}$
$B=$ $\frac{x+8\sqrt{x}+16}{(\sqrt{x}-3)(\sqrt{x}+4)}$
a. Rút gọn A;B
b. Tính B khi $x=\frac{\sqrt{6}}{\sqrt{1,5}}$
c. Rút gọn $Q=\frac{A}{B}$
d. Tìm min $Q$
e. Tìm $x∈Z$ để $Q∈Z$
Đáp án:
d. \(Min = – \dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
A = \dfrac{{\sqrt x – 3 – 5 + \left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x – 8 + x – 4}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + \sqrt x – 12}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 2} \right)}} = \dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}\\
DK:x \ge 0;x \ne 9\\
B = \dfrac{{{{\left( {\sqrt x + 4} \right)}^2}}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 4} \right)}} = \dfrac{{\sqrt x + 4}}{{\sqrt x – 3}}\\
b.Thay:x = \sqrt {\dfrac{6}{{1,5}}} = 2\\
\to B = \dfrac{{\sqrt 2 + 4}}{{\sqrt 2 + 2}} = 3 – \sqrt 2 \\
c.Q = \dfrac{A}{B} = \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}:\dfrac{{\sqrt x + 4}}{{\sqrt x – 3}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}.\dfrac{{\sqrt x – 3}}{{\sqrt x + 4}} = \dfrac{{\sqrt x – 3}}{{\sqrt x + 2}}\\
d.Q = \dfrac{{\sqrt x – 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 – 5}}{{\sqrt x + 2}} = 1 – \dfrac{5}{{\sqrt x + 2}}\\
Do:x \ge 0 \to \sqrt x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{5}{{\sqrt x + 2}} \le \dfrac{5}{2}\\
\to – \dfrac{5}{{\sqrt x + 2}} \ge – \dfrac{5}{2}\\
\to 1 – \dfrac{5}{{\sqrt x + 2}} \ge – \dfrac{3}{2}\\
\to Min = – \dfrac{3}{2}\\
\Leftrightarrow x = 0\\
e.Q \in Z\\
\Leftrightarrow \dfrac{5}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 \in U\left( 5 \right)\\
Mà:\sqrt x + 2 \ge 2\forall x \ge 0\\
\to \sqrt x + 2 = 5\\
\to \sqrt x = 3\\
\to x = 9\left( l \right)\\
\to x \in \emptyset
\end{array}\)
Đáp án:
Giải thích các bước giải: