18.
\(y=a\) là tiệm cận ngang \(\Rightarrow a=-1\)
\(x=-c\) là tiệm cận đứng \(\Rightarrow c=-1\)
\(\Rightarrow y=\dfrac{-x+b}{x-1}\)
Đồ thị hàm số đi qua điểm \(\left(2;0\right)\Rightarrow\dfrac{-2+b}{2-1}=0\Rightarrow b=2\)
\(\Rightarrow T=0\)
19.
\(P=\dfrac{a^{\sqrt{2022}+1+2-\sqrt{2022}}}{a^{\left(\sqrt{2}-2\right)\left(\sqrt{2}+2\right)}}=\dfrac{a^3}{a^{-2}}=a^5\)
20.
\(T=2(a+b)^{-1}.(ab)^{\frac{1}{2}}\left[1+\dfrac{1}{4}\left(\sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}} \right)^2 \right]^\frac{1}{2}\)
\(=2(a+b)^{-1}(ab)^{\frac{1}{2}}\)\(\left[1+\dfrac{1}{4}.\dfrac{\left(a-b\right)^2}{ab}\right]^{\dfrac{1}{2}}\)
\(=2(a+b)^{-1}(ab)^{\frac{1}{2}}\)\(\left[\dfrac{a^2+b^2+2ab}{4ab}\right]^{\dfrac{1}{2}}\)
\(=2(a+b)^{-1}(ab)^{\frac{1}{2}}.\dfrac{a+b}{2(ab)^{\frac{1}{2}}}\)
\(=1\)
21.
Do số mũ \(\dfrac{1}{3}\) không nguyên nên:
ĐKXĐ: \(3x^2-1>0\Rightarrow x\in\left(-\infty;-\dfrac{1}{\sqrt{3}}\right)\cup\left(\dfrac{1}{\sqrt{3}};+\infty\right)\)