a.Thế \(x=2\) vào \(B\) ta được:
\(B=\dfrac{-10}{2-4}=\dfrac{-10}{-2}=5\)
b.\(A=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+6x+5}-\dfrac{1}{1+x}\)
\(A=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{\left(x+1\right)\left(x+5\right)}-\dfrac{1}{1+x}\)
\(A=\dfrac{\left(x+2\right)\left(x+1\right)-5x-1-\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}\)
\(A=\dfrac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)
\(A=\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\)
\(A=\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\)
\(A=\dfrac{x-4}{x+5}\)
c.\(P=A.B=\dfrac{x-4}{x+5}.\dfrac{-10}{x-4}=\dfrac{-10\left(x-4\right)}{\left(x+5\right)\left(x-4\right)}=-\dfrac{10}{x+5}\)
Để P nhận giá trị nguyên thì \(-\dfrac{10}{x+5}\in Z\) hay \(-x-5\in U\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
-x-5=1 => x=-6
-x-5=-1 => x=-4
-x-5=2 => x=-7
-x-5=-2 => x=-3
-x-5=5 => x=-10
-x-5=-5 => x=0
-x-5=10 => x=-15
-x-5=-10 => x=5
Vậy \(x\in\left\{-6;-4;-7;-3;-10;0;-15;5\right\}\) thì P nhận giá trị nguyên
a thay x=2 vào biểu thức B ta có:
-10/-2-4
=-10/-6
=5/3
