Cho biểu thức A=(x/x²-9+3/3-x+1/x+3)÷(x-3+10-x²/x+3). Rút gọn rồi tìm giá trị của x để A<0. 20/07/2021 Bởi Iris Cho biểu thức A=(x/x²-9+3/3-x+1/x+3)÷(x-3+10-x²/x+3). Rút gọn rồi tìm giá trị của x để A<0.
Đáp án: $\begin{array}{l}A = \left( {\frac{x}{{{x^2} – 9}} + \frac{3}{{3 – x}} + \frac{1}{{x + 3}}} \right):\left( {x – 3 + \frac{{10 – {x^2}}}{{x + 3}}} \right)\\Dkxd:x \ne 3;x \ne – 3\\A = \left( {\frac{x}{{\left( {x – 3} \right)\left( {x + 3} \right)}} – \frac{3}{{x – 3}} + \frac{1}{{x + 3}}} \right):\frac{{\left( {x – 3} \right)\left( {x + 3} \right) + 10 – {x^2}}}{{x + 3}}\\ = \frac{{x – 3\left( {x + 3} \right) + \left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{{x^2} – 9 + 10 – {x^2}}}\\ = \frac{{x – 3x – 9 + x – 3}}{{x – 3}}.\frac{1}{1}\\ = \frac{{ – x – 12}}{{x – 3}}\\A < 0\\ \Rightarrow \frac{{ – x – 12}}{{x – 3}} < 0\\ \Rightarrow \frac{{x + 12}}{{x – 3}} > 0\\ \Rightarrow \left[ \begin{array}{l}x > 3\\x < – 12\end{array} \right.\end{array}$ Vậy x>3 hoặc x<-12 thì A<0 Bình luận
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Đáp án:
$\begin{array}{l}
A = \left( {\frac{x}{{{x^2} – 9}} + \frac{3}{{3 – x}} + \frac{1}{{x + 3}}} \right):\left( {x – 3 + \frac{{10 – {x^2}}}{{x + 3}}} \right)\\
Dkxd:x \ne 3;x \ne – 3\\
A = \left( {\frac{x}{{\left( {x – 3} \right)\left( {x + 3} \right)}} – \frac{3}{{x – 3}} + \frac{1}{{x + 3}}} \right):\frac{{\left( {x – 3} \right)\left( {x + 3} \right) + 10 – {x^2}}}{{x + 3}}\\
= \frac{{x – 3\left( {x + 3} \right) + \left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{{x^2} – 9 + 10 – {x^2}}}\\
= \frac{{x – 3x – 9 + x – 3}}{{x – 3}}.\frac{1}{1}\\
= \frac{{ – x – 12}}{{x – 3}}\\
A < 0\\
\Rightarrow \frac{{ – x – 12}}{{x – 3}} < 0\\
\Rightarrow \frac{{x + 12}}{{x – 3}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 3\\
x < – 12
\end{array} \right.
\end{array}$
Vậy x>3 hoặc x<-12 thì A<0