a) B=\(\left(\dfrac{9-3x}{x^2+4x-5}-\dfrac{x+5}{1-x}-\dfrac{x+1}{x+5}\right):\dfrac{7x-14}{x^3-1}\)
= \(\left(\dfrac{9-3x}{\left(x-1\right)\left(x+5\right)}+\dfrac{x+5}{x-1}-\dfrac{x+1}{x+5}\right):\dfrac{7x-14}{x^3-1}\)
= \(\left(\dfrac{9-3x}{\left(x-1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\dfrac{x^2-1}{\left(x-1\right)\left(x+5\right)}\right):\dfrac{7x-14}{x^3-1}\)
=\(\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}:\dfrac{7x-14}{x^3-1}\)
=\(\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{7x-14}\)
= \(\dfrac{7\left(x+5\right)}{x+5}.\dfrac{x^2+x+1}{7\left(x-2\right)}\)
= \(\dfrac{x^2+x+1}{x-2}\) => đpcm
b) (x + 5)2 - 9x - 45 =0
⇔ x2 + 10x +25 - 9x - 45 =0
⇔ x2 + x - 20 = 0
⇔ (x-4)(x+5)=0
⇔ \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-5\left(loại\right)\end{matrix}\right.\)
B=\(\dfrac{4^2+4+1}{4-2}=\dfrac{21}{2}\)
