a: \(B=\left(\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right):\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{3x^2+11x+10}\)
\(=\dfrac{3x^2+5x-\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)\left(3x+5\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\dfrac{3x^2+5x-x^2-2x+x+2}{x-3}\cdot\dfrac{1}{\left(x-1\right)^2\cdot\left(x+1\right)^2}\)
\(=\dfrac{2x^2-4x+2}{\left(x-3\right)}\cdot\dfrac{1}{\left(x-1\right)^2\cdot\left(x+1\right)^2}=\dfrac{2}{\left(x-3\right)\left(x+1\right)^2}\)
c: Khi x>3 thì (x-3)>0
\(\Leftrightarrow B>0\)
a. \(B=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\left(x^2-2x+1\right)\)
\(=\left(\dfrac{x}{x^2+2x-3x-6}-\dfrac{x-1}{3x^2-9x+5x-15}\right):\dfrac{\left(x^2-1\right)^2\left(x-1\right)^2}{3x^2+6x+5x+10}\)
\(=\left[\dfrac{x}{x\left(x+2\right)-3\left(x+2\right)}-\dfrac{x-1}{3x\left(x-3\right)+5\left(x-3\right)}\right]:\dfrac{\left[\left(x+1\right)\left(x-1\right)\right]^2\left(x-1\right)^2}{3x\left(x+2\right)+5\left(x+2\right)}\)
\(=\left[\dfrac{x}{\left(x+2\right)\left(x-3\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x+1\right)^2\left(x-1\right)^2\left(x-1\right)^2}{\left(x+2\right)\left(3x+5\right)}\)
\(=\dfrac{x\left(3x+5\right)-\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-3\right)\left(3x+5\right)}:\dfrac{\left(x+1\right)^2\left(x-1\right)^4}{\left(x+2\right)\left(3x+5\right)}\)
\(=\dfrac{2x^2+4x+2}{\left(x+2\right)\left(x-3\right)\left(3x+5\right)}:\dfrac{\left(x+1\right)^2\left(x-1\right)^4}{\left(x+2\right)\left(3x+5\right)}\)
\(=\dfrac{2\left(x^2+2x+1\right)}{\left(x+2\right)\left(x-3\right)\left(3x+5\right)}:\dfrac{\left(x+1\right)^2\left(x-1\right)^4}{\left(x+2\right)\left(3x+5\right)}\)
\(=\dfrac{2\left(x+1\right)^2}{\left(x+2\right)\left(x-3\right)\left(3x+5\right)}:\dfrac{\left(x+1\right)^2\left(x-1\right)^4}{\left(x+2\right)\left(3x+5\right)}\)
\(=\dfrac{2\left(x+1\right)^2\left(x+2\right)\left(3x+5\right)}{\left(x+2\right)\left(x-3\right)\left(3x+5\right)\left(x+1\right)^2\left(x-1\right)^4}\)
\(=\dfrac{2}{\left(x-3\right)\left(x-1\right)^4}\)
Mình đang nghĩ câu b và c nhé.
