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”
Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l} A = \left( {\frac{{\sqrt x }}{{\sqrt x + 3}} – \frac{{x + 9}}{{x – 9}}} \right):\left( {\frac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \frac{1}{{\sqrt x }}} \right)\\ = \left( {\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}} – \frac{{x + 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\left( {\frac{{3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 3} \right)}} – \frac{{\sqrt x – 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right)\\ = \left( {\frac{{x – 3\sqrt x – x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\left( {\frac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right)\\ = \frac{{ – 3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}:\frac{{2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 3} \right)}}\\ = \frac{{ – 3}}{{\sqrt x – 3}}.\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)}}\\ = \frac{{ – 3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}} \end{array}\)
b,
\[\begin{array}{l} A > – 1 \Leftrightarrow \frac{{ – 3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}} > – 1\\ \Leftrightarrow \frac{{3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}} < 1\\ \Leftrightarrow 3\sqrt x < 2\sqrt x + 4\\ \Leftrightarrow \sqrt x < 4\\ \Leftrightarrow x < 16 \end{array}\]
Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
A = \left( {\frac{{\sqrt x }}{{\sqrt x + 3}} – \frac{{x + 9}}{{x – 9}}} \right):\left( {\frac{{3\sqrt x + 1}}{{x – 3\sqrt x }} – \frac{1}{{\sqrt x }}} \right)\\
= \left( {\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}} – \frac{{x + 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\left( {\frac{{3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 3} \right)}} – \frac{{\sqrt x – 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right)\\
= \left( {\frac{{x – 3\sqrt x – x – 9}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\left( {\frac{{3\sqrt x + 1 – \sqrt x + 3}}{{\sqrt x \left( {\sqrt x – 3} \right)}}} \right)\\
= \frac{{ – 3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}:\frac{{2\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 3} \right)}}\\
= \frac{{ – 3}}{{\sqrt x – 3}}.\frac{{\sqrt x \left( {\sqrt x – 3} \right)}}{{2\left( {\sqrt x + 2} \right)}}\\
= \frac{{ – 3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}}
\end{array}\)
b,
\[\begin{array}{l}
A > – 1 \Leftrightarrow \frac{{ – 3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}} > – 1\\
\Leftrightarrow \frac{{3\sqrt x }}{{2\left( {\sqrt x + 2} \right)}} < 1\\
\Leftrightarrow 3\sqrt x < 2\sqrt x + 4\\
\Leftrightarrow \sqrt x < 4\\
\Leftrightarrow x < 16
\end{array}\]