a:
Sửa đề; \(D=\dfrac{2x-6}{x^3-3x^2+5x-15}\)
ĐKXĐ: x<>3
\(D=\dfrac{2\left(x-3\right)}{x^2\left(x-3\right)+5\left(x-3\right)}=\dfrac{2}{x^2+5}\)
b: Để D=1/5 thì x^2+5=10
=>x^2=5
hay \(x=\pm\sqrt{5}\)
c: Để D=1/x^2+1 thì \(\dfrac{2}{x^2+5}=\dfrac{1}{x^2+1}\)
=>\(2x^2+2=x^2+5\)
=>x^2=3
hay \(x=\pm\sqrt{3}\)