Đáp án:
$B$
Giải thích các bước giải:
$y=\dfrac{x-1}{\sqrt{x^2+2(m-1)x+m^2}}\\ f(x)=x^2+2(m-1)x+m^2\\ \Delta'=(m-1)^2-m^2=-2m+1<0 \ \forall m >\dfrac{1}{2}\\ \Rightarrow x^2+2(m-1)x+m^2 >0 \ \forall x; \forall m>\dfrac{1}{2}(\text{Do hệ số góc a=1>0})\\ \Rightarrow \dfrac{x-1}{\sqrt{x^2+2(m-1)x+m^2}} \text{ xác định với} \ \forall x; \forall m>\dfrac{1}{2}\\ \displaystyle \lim_{x \to -\infty} \dfrac{x-1}{\sqrt{x^2+2(m-1)x+m^2}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{x-1}{|x|\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{x-1}{-x\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{1-\dfrac{1}{x}}{-\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\dfrac{1}{-1}\\ =-1\\ \displaystyle \lim_{x \to +\infty} \dfrac{x-1}{\sqrt{x^2+2(m-1)x+m^2}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{x-1}{|x|\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{x-1}{x\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{1-\dfrac{1}{x}}{\sqrt{1+\dfrac{2(m-1)}{x}+\dfrac{m^2}{x^2}}}\\ =\dfrac{1}{1}\\ =1$
$f(x)=x^2+2(m-1)x+m^2$ vô nghiệm nên $y$ không có TCĐ.
