a) ĐKXĐ: \(x\ne3;-2;-\dfrac{5}{3}\)
\(B=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\left(x^2-2x+1\right)\)
\(=\left(\dfrac{x}{\left(x+2\right)\left(x-3\right)}-\dfrac{x-1}{\left(3x+5\right)\left(x-3\right)}\right):\dfrac{\left(x^2-1\right)^2}{\left(x+2\right)\left(3x+5\right)}\left(x-1\right)^2\)
\(=\dfrac{x\left(3x+5\right)-\left(x-1\right)\left(x+2\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x-1\right)^2\left(x+1\right)^2}.\left(x-1\right)^2\)
\(=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x+1\right)^2}\)
\(=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}.\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x+1\right)^2}=\dfrac{2}{x-3}\)
b) \(B=\dfrac{2}{2003-3}=\dfrac{2}{2000}=\dfrac{1}{1000}\)
c) \(x>3\Rightarrow x-3>0\Rightarrow\dfrac{2}{x-3}>0\)
a) ĐKXĐ: \(x\notin\left\{3;-\dfrac{3}{5};-2;-1;1\right\}\)
Ta có: \(B=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\)
\(=\left(\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right):\dfrac{\left(x^2-1\right)^2}{3x^2+6x+5x+10}\cdot\left(x-1\right)^2\)
\(=\dfrac{x\left(3x+5\right)-\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)\left(3x+5\right)}:\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{3x\left(x+2\right)+5\left(x+2\right)}\cdot\left(x-1\right)^2\)
\(=\dfrac{3x^2+5x-x^2-x+2}{\left(x-3\right)\left(x+2\right)\left(3x+5\right)}\cdot\dfrac{\left(x+2\right)\left(3x+5\right)}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2}{1}\)
\(=\dfrac{2x^2+4x+2}{x-3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\dfrac{2\left(x^2+2x+1\right)}{x-3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\dfrac{2}{x-3}\)
b) Thay x=2003 vào B, ta được:
\(B=\dfrac{2}{2003-3}=\dfrac{2}{2000}=\dfrac{1}{1000}\)
c) Để B>0 thì x-3>0
hay x>3
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