cho a=(căn x/căn x+3-x+9/x-9):(3^x+1/x-3căn x-1/căn x) (x>0;x#9) A.rut gọn biểu thức a B.tim x sao cho a>-1
0 bình luận về “cho a=(căn x/căn x+3-x+9/x-9):(3^x+1/x-3căn x-1/căn x) (x>0;x#9) A.rut gọn biểu thức a B.tim x sao cho a>-1”
Đáp án:
$\begin{array}{l} A = \left( {\frac{{\sqrt x }}{{\sqrt x – 3}} – \frac{{x + 9}}{{x – 9}}} \right):\left( {\frac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \frac{1}{{\sqrt x }}} \right)\left( {x > 0;x \ne 9} \right)\\ = \frac{{\sqrt x \left( {\sqrt x + 3} \right) – \left( {x + 9} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\left[ {\frac{{3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 3} \right)}} – \frac{1}{{\sqrt x }}} \right]\\ = \frac{{x + 3\sqrt x – x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\frac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}\\ = \frac{{3\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\sqrt x + 4}}\\ = \frac{{3\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}} \end{array}$
Đáp án:
$\begin{array}{l}
A = \left( {\frac{{\sqrt x }}{{\sqrt x – 3}} – \frac{{x + 9}}{{x – 9}}} \right):\left( {\frac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \frac{1}{{\sqrt x }}} \right)\left( {x > 0;x \ne 9} \right)\\
= \frac{{\sqrt x \left( {\sqrt x + 3} \right) – \left( {x + 9} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\left[ {\frac{{3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 3} \right)}} – \frac{1}{{\sqrt x }}} \right]\\
= \frac{{x + 3\sqrt x – x – 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}:\frac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}\\
= \frac{{3\left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right)}}.\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\sqrt x + 4}}\\
= \frac{{3\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)\left( {\sqrt x + 3} \right)}}
\end{array}$