\(A=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+6x+5}-\dfrac{1}{1+x}\)
\(=\dfrac{x+2}{x+5}-\dfrac{5x+1}{\left(x+1\right)\left(x+5\right)}-\dfrac{1}{x+1}\)
\(=\dfrac{\left(x+2\right)\left(x+1\right)-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}=\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{x-4}{x+5}\)
b) \(B=A.\dfrac{-10}{x-4}=\dfrac{x-4}{x+5}.\dfrac{-10}{x-4}=-\dfrac{10}{x+5}\)
\(A=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+6x+5}-\dfrac{1}{1+x}\)
=\(\dfrac{(x+2)\left(x+1\right)}{(x+5)\left(x+1\right)}-\dfrac{5x+1}{\left(x+1\right)\left(x+5\right)}-\dfrac{x+5}{\left(x+5\right)\left(x+1\right)}\)
=\(\dfrac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}=\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)
=\(\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{x-4}{x+5}\)
b, B=A.\(\dfrac{-10}{x-4}=\dfrac{x-4}{x+5}.\dfrac{-10}{x-4}=\dfrac{-10}{x+5}\)
để B nguyên khi \(\dfrac{-10}{x+5}\) nguyên
=>\(\dfrac{-10}{x+5}\inƯ\left(10\right)=\left\{-1;1;-2;2;-5;5;-10;10\right\}\)
với x+5=-1=>x=-6( thỏa mãn)
với x+5=1=>x=-4(thỏa mãn)
với x+5=-2=>x=-7(thỏa mãn)
với x+5=2=>x=-3(thỏa mãn)
với x+5=-5=>x=-10(thỏa mãn)
với x+5=5=>x=0(thỏa mãn)
với x+5=-10=>x=-15(thỏa mãn)
với x+5=10=>x=5(thoa mãn)
vậy x=(-6;-4;-7;-3;-10;0;-15;5) thì B nguyên
