Question

P= \(\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\)(x≠2;x≠3)

Answer 1

\(P=\dfrac{2x-9-\left(x+3\right)\left(x-3\right)+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)

Answer 2

\(P=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x+3}{x-2}-\dfrac{2x+1}{3-x}\left(x\ne2;x\ne3\right)\)

\(=\dfrac{2x-9}{x^2-2x-3x+6}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{2x+1}{x-3}\)

\(=\dfrac{2x-9}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}+\dfrac{2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2-2x+x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x\left(x-2\right)+\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+1}{x-3}\)

#Urushi