\(y'=\dfrac{-m^2-5}{\left(x-m\right)^2}< 0\) \(\Rightarrow\) hàm nghịch biến trên các khoảng xác định

\(\Rightarrow\) Hàm đạt min trên [0;1] bằng -7 khi:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}m>1\\m< 0\end{matrix}\right.\\f\left(1\right)=-7\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>1\\m< 0\end{matrix}\right.\\\dfrac{m+5}{1-m}=-7\end{matrix}\right.\) \(\Rightarrow m=2\)

17.

\(log_a\dfrac{a}{bc}=log_aa-log_a\left(bc\right)=log_aa-\left(log_ab+lo_ac\right)\)

\(=1-\left(2+3\right)=-4\)

18.

\(log_a\left(a^3b^2\sqrt{c}\right)=log_aa^3+log_ab^2+log_a\sqrt{c}\)

\(=3log_aa+2log_ab+\dfrac{1}{2}log_ac=3+2.3+\dfrac{1}{2}.\left(-1\right)=8\)

19.

\(log_{27}\left(\dfrac{\sqrt{x}}{y}\right)^3=3log_{3^3}\left(\dfrac{\sqrt{x}}{y}\right)=3.\dfrac{1}{3}log_3\left(\dfrac{\sqrt{x}}{y}\right)\)

\(=log_3\sqrt{x}-log_3y=\dfrac{1}{2}log_3x-log_3y=\dfrac{\alpha}{2}-\beta\)

20.

\(\log_{\sqrt{a}}(a^2\sqrt{b})=\log_{a^\frac{1}{2}}(a^2\sqrt{b})=2\log_a(a^2\sqrt{b})\)

\(=2\log_aa^2+2\log_a\sqrt{b}=4\log_aa+\log_ab=4+\log_ab\)

21.

\(y'=\dfrac{\left(2+2019^x\right)'}{\left(2+2019^x\right).\ln2019}=\dfrac{2019^x.\ln2019}{\left(2+2019^x\right).\ln2019}=\dfrac{2019^x}{2+2019^x}\)

22.

Do \(\dfrac{\sqrt{3}}{3}< \dfrac{\sqrt{2}}{2}\) mà \(a^{\frac{\sqrt{3}}{3}}>a^{\frac{\sqrt{2}}{2}}\) \(\Rightarrow0< a< 1\)

Do \(\dfrac{3}{4}< \dfrac{4}{5}\) mà \(\log_b\dfrac{3}{4}< \log_b\dfrac{4}{5}\Rightarrow b>1\)

Vậy \(0< a< 1,b>1\)

23.

Hàm đồng biến trên R khi:

\(a^2-3a+3>1\Leftrightarrow a^2-3a+2>0\Rightarrow\left[{}\begin{matrix}a< 1\\a>2\end{matrix}\right.\)

\(f\left(1\right)-f\left(0\right)=\int_0^1\dfrac{1}{\sqrt{x+1}-\sqrt{x}}dx\)

\(=\int_0^1\left(\sqrt{x+1}+\sqrt{x}\right)dx\)

\(=\left(\dfrac{2}{3}.\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}.\left(x\right)^{\dfrac{3}{2}}\right)|_0^1\)

\(=\dfrac{2}{3}\left(2^{\dfrac{3}{2}}+1\right)-\dfrac{2}{3}\)

\(=\dfrac{2}{3}\left(\sqrt{8}+1\right)-\dfrac{2}{3}=\dfrac{4\sqrt{2}}{3}\)