Giúp em vs ạ em cần nộp gấp ai đầu em vote 5 sao ạ C = √x + 2/√x + 3 + √x/ 2 - √x + x + 2√x - 8/(√x + 3)(√x -2) a,tìm điều kiện xác định b,rút gọn C c, tìm x để c<1
1 câu trả lời
Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x - 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
b)\\
C = \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{2 - \sqrt x }} + \dfrac{{x + 2\sqrt x - 8}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \sqrt x \left( {\sqrt x + 3} \right) + x + 2\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 4 - x - 3\sqrt x + x + 2\sqrt x - 8}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - \sqrt x - 12}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}\\
c)C < 1\\
\Leftrightarrow \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}} - 1 < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 4 - \left( {\sqrt x - 2} \right)}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x - 4 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \dfrac{{ - 2}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \sqrt x - 2 > 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4\\
Vậy\,x > 4
\end{array}$