Đáp án:
\(\begin{array}{l}
B5:\\
a)A = \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}}\\
b)A = - \dfrac{1}{4}\\
c)0 \le x < 25\\
d)Min = - \dfrac{{25}}{4}\\
B6:\\
a)A = x - 1\\
b)0 < x < 1\\
c)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a)DK:x \ge 0;x \ne 25\\
A = \dfrac{{\sqrt x \left( {\sqrt x + 5} \right) - 10\sqrt x - 5\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x + 5\sqrt x - 10\sqrt x - 5\sqrt x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{x - 10\sqrt x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 5} \right)}^2}}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}}\\
b)Thay:x = 9\\
\to A = \dfrac{{\sqrt 9 - 5}}{{\sqrt 9 + 5}} = \dfrac{{ - 2}}{8} = - \dfrac{1}{4}\\
c)A < \dfrac{1}{3}\\
\to \dfrac{{\sqrt x - 5}}{{\sqrt x + 5}} < \dfrac{1}{3}\\
\to \dfrac{{3\sqrt x - 15}}{{3\left( {\sqrt x + 5} \right)}} < 0\\
\to 3\sqrt x - 15 < 0\left( {do:\sqrt x + 5 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 25\\
d)B = \left( {x + 5\sqrt x } \right).A\\
= \left( {x + 5\sqrt x } \right).\dfrac{{\sqrt x - 5}}{{\sqrt x + 5}}\\
= \sqrt x \left( {\sqrt x - 5} \right)\\
= x - 5\sqrt x \\
= x - 2.\sqrt x .\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{25}}{4}\\
= {\left( {\sqrt x - \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4}\\
Do:{\left( {\sqrt x - \dfrac{5}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} \ge - \dfrac{{25}}{4}\\
\to Min = - \dfrac{{25}}{4}\\
\Leftrightarrow \sqrt x - \dfrac{5}{2} = 0\\
\to x = \dfrac{{25}}{4}\\
B6:\\
a)DK:x > 0;x \ne 1\\
A = \dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\sqrt x \left( {\sqrt x - 1} \right)\\
= x - 1\\
b)A < 0\\
\to x - 1 < 0\\
\to 0 < x < 1\\
c)A = 2\\
\to x - 1 = 2\\
\to x = 3\\
d)B = A + \dfrac{5}{{\sqrt x }}\\
= x + 1 + \dfrac{5}{{\sqrt x }}
\end{array}\)
( bài 6 bạn xem lại đề nha, rút gọn xong làm đến câu d đề hơi sai nha bạn )
