giúp mik câu này thật nhanh nha
a) Để A xác định thì: \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\\\dfrac{x-1}{x+3}\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne1\end{matrix}\right.\)
Với \(x\ne\pm3;x\ne1\), ta có:
\(A=\left(\dfrac{2x}{x-3}+\dfrac{x}{x+3}+\dfrac{2x^2+3x+1}{9-x^2}\right):\dfrac{x-1}{x+3}\)
\(=\left[\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+3x+1}{\left(x-3\right)\left(x+3\right)}\right]\cdot\dfrac{x+3}{x-1}\)
\(=\dfrac{2x^2+6x+x^2-3x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-1}\)
\(=\dfrac{x^2-1}{x-3}\cdot\dfrac{1}{x-1}\)
\(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-3\right)\left(x-1\right)}=\dfrac{x+1}{x-3}\)
Vậy \(A=\dfrac{x+1}{x-3}\) với \(x\ne\pm3;x\ne1\).
b) Với \(x\ne\pm3;x\ne1\):
Để \(A=3\) thì \(\dfrac{x+1}{x-3}=3\)
\(\Rightarrow x+1=3\left(x-3\right)\)
\(\Leftrightarrow x+1=3x-9\)
\(\Leftrightarrow x-3x=-9-1\)
\(\Leftrightarrow-2x=-10\Leftrightarrow x=5\) (tm ĐKXĐ)
Vậy \(A=3\) tại \(x=5\).
c) Để \(A< 1\) thì \(\dfrac{x+1}{x-3}< 1\)
\(\Leftrightarrow\dfrac{x+1}{x-3}-1< 0\)
\(\Leftrightarrow\dfrac{x+1-\left(x-3\right)}{x-3}< 0\)
\(\Leftrightarrow\dfrac{4}{x-3}< 0\)
\(\Rightarrow x-3< 0\) (vì \(4>0\))
\(\Leftrightarrow x< 3\)
Kết hợp với ĐKXĐ của \(x\), ta được: \(x< 3;x\ne-3;x\ne1\)
Vậy \(A< 1\) khi \(x< 3;x\ne-3;x\ne1\).
\(\text{#}Toru\)
