Đáp án:
B13:
c. x=16
Giải thích các bước giải:
\(\begin{array}{l}
B13:\\
a.DK:x \ne 4;x \ge 0\\
b.P = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - \left( {2 + 5\sqrt x } \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
c.P = 2\\
\to \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = 2\\
\to 3\sqrt x = 2\sqrt x + 4\\
\to \sqrt x = 4\\
\to x = 16\left( {TM} \right)\\
B15:\\
a,DK:x \ge 0;x \ne 1\\
b.K = \dfrac{{15\sqrt x - 11 - 3\sqrt x \left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 9\sqrt x - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 5\sqrt x - 8}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
c.K = \dfrac{1}{2}\\
\to \dfrac{{ - 5x + 5\sqrt x - 8}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}} = \dfrac{1}{2}\\
\to - 10x + 10\sqrt x - 16 = x + 2\sqrt x - 3\\
\to 11x - 8\sqrt x + 13 = 0\left( {vô nghiệm} \right)\\
\to x \in \emptyset
\end{array}\)
