Đáp án:

$32)$Hàm số có 3 tiệm cận

$35)y=\dfrac{5}{2}.$

Giải thích các bước giải:

$32)\\ y=\dfrac{x-2}{\sqrt{x^2-5x+6}} \ \ \ \ D= (-\infty;2) \cup (3;+\infty)\\ =\dfrac{x-2}{\sqrt{(x-2)(x-3)}}\\ \displaystyle \lim_{x \to 2^+}\dfrac{x-2}{\sqrt{(x-2)(x-3)}}=\displaystyle \lim_{x \to 2^-}\dfrac{x-2}{\sqrt{(x-2)(x-3)}}=\displaystyle \lim_{x \to 2^-}\dfrac{\sqrt{x-2}}{\sqrt{x-3}}=0\\ \displaystyle \lim_{x \to 3^+}\dfrac{x-2}{\sqrt{(x-2)(x-3)}}=+\infty\\ \Rightarrow \text{TCĐ:  }x=3\\ \displaystyle \lim_{x \to +\infty} \dfrac{x-2}{\sqrt{x^2-5x+6}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{x-2}{|x|\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{x-2}{x\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{1-\dfrac{2}{x}}{\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =1\\ \displaystyle \lim_{x \to -\infty} \dfrac{x-2}{\sqrt{x^2-5x+6}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{x-2}{|x|\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{x-2}{-x\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{1-\dfrac{2}{x}}{-\sqrt{1-\dfrac{5}{x}+\dfrac{6}{x^2}}}\\ =-1\\ \Rightarrow \text{TCN:  }y= \pm 1$

Hàm số có 3 tiệm cận

$35)\\ y=x-\sqrt{x^2-5x} \ \ \ \ D=(-\infty;0] \cup [5;+\infty)\\ \displaystyle \lim_{x \to +\infty} (x-\sqrt{x^2-5x})\\ =\displaystyle \lim_{x \to +\infty} \dfrac{(x-\sqrt{x^2-5x})(x+\sqrt{x^2-5x})}{x+\sqrt{x^2-5x}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{5x}{x+\sqrt{x^2-5x}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{5x}{x+|x|\sqrt{1-\dfrac{5}{x}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{5x}{x+x\sqrt{1-\dfrac{5}{x}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{5}{1+1\sqrt{1-\dfrac{5}{x}}}\\ =\dfrac{5}{2}\\ \displaystyle \lim_{x \to -\infty} (x-\sqrt{x^2-5x})\\ =\displaystyle \lim_{x \to -\infty} \dfrac{(x-\sqrt{x^2-5x})(x+\sqrt{x^2-5x})}{x+\sqrt{x^2-5x}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{5x}{x+\sqrt{x^2-5x}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{5x}{x+|x|\sqrt{1-\dfrac{5}{x}}}\\ =\displaystyle \lim_{x \to -\infty} \dfrac{5x}{x-x\sqrt{1-\dfrac{5}{x}}}\\ =\displaystyle \lim_{x \to +\infty} \dfrac{5}{1-1\sqrt{1-\dfrac{5}{x}}}\\ = -\infty\\ \Rightarrow \text{TCN: }y=\dfrac{5}{2}.$