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a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b)
ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)
Ta có: P=AB
\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)
\(=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)
\(\Leftrightarrow9\left(x+1\right)=6x\)
\(\Leftrightarrow9x-6x=-9\)
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)
\(a,P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\left(x\ne\pm3;x\ne-2\right)\\ P=\dfrac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\\ P=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}=\dfrac{3x}{x-3}\\ b,x^2-7x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow x=4\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{3\cdot4}{4-3}=12\\ c,P=\dfrac{3\left(x-3\right)+9}{x-3}=3+\dfrac{9}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(9\right)=\left\{-9;-3;-1;1;3;9\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;4;6;12\right\}\)
a) \(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
\(\Leftrightarrow A=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{2x^2-6x}{\left(x+3\right)\left(x-3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{3x^2-13x}{x^2-9}\)
\(A=\frac{2x}{x+3}-\frac{x+1}{3-x}-\frac{3-11x}{x^2-9}\)
a) ĐK : x ≠ ±3
\(=\frac{2x}{x+3}+\frac{x+1}{x-3}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x}{\left(x-3\right)\left(x+3\right)}+\frac{x^2+4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3x}{x-3}\)
b) Để A < 2
=> \(\frac{3x}{x-3}< 2\)
<=> \(\frac{3x}{x-3}-2< 0\)
<=> \(\frac{3x}{x-3}-\frac{2x-6}{x-3}< 0\)
<=> \(\frac{3x-2x+6}{x-3}< 0\)
<=> \(\frac{x+6}{x-3}< 0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}x+6>0\\x-3< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-6\\x< 3\end{cases}}\Leftrightarrow-6< x< 3\)
2. \(\hept{\begin{cases}x+6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< -6\\x>3\end{cases}}\)( loại )
Vậy -6 < x < 3
1: \(B=\dfrac{6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x^2+3x}{\left(x+3\right)\left(x-3\right)}=\dfrac{x}{x-3}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}\)
\(\Leftrightarrow A=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2\left(x+2\right)}{x-3}\)
\(\Leftrightarrow A=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow A=\frac{x+4}{x-3}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow\frac{x+4}{x-3}\inℤ\)
\(\Leftrightarrow1+\frac{7}{x-3}\inℤ\)
\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;4;-4;10\right\}\)
c) Để \(A=\frac{3}{5}\)
\(\Leftrightarrow\frac{x+4}{x-3}=\frac{3}{5}\)
\(\Leftrightarrow5x+20=3x-9\)
\(\Leftrightarrow2x+29=0\)
\(\Leftrightarrow x=-\frac{29}{2}\)
d) Để \(A< 0\)
\(\Leftrightarrow\frac{x+4}{x-3}< 0\)
\(\Leftrightarrow1+\frac{7}{x-3}< 0\)
\(\Leftrightarrow\frac{-7}{x-3}< 1\)
\(\Leftrightarrow-7< x-3\)
\(\Leftrightarrow x>-4\)
e) Để \(A>0\)
\(\Leftrightarrow\frac{x+4}{x-3}>0\)
\(\Leftrightarrow1+\frac{7}{x-3}>0\)
\(\Leftrightarrow\frac{-7}{x-3}>1\)
\(\Leftrightarrow-7>x-3\)
\(\Leftrightarrow x< -4\)