Bài 3: Lôgarit

ĐKXĐ:
1.

\(\left\{{}\begin{matrix}x-1>0\\log_2\left(x-1\right)-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x-1>2\end{matrix}\right.\) \(\Leftrightarrow x>3\)

2.

\(\left\{{}\begin{matrix}x-1>0\\x-1\ne1\\\dfrac{x}{2-x^2}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x\ne2\\\left[{}\begin{matrix}x< -\sqrt{2}\\0< x< \sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow1< x< \sqrt{2}\)

Đặt \(2^x=t>0\) ta được:

\(t^2-5t+4=0\Rightarrow\left[{}\begin{matrix}t=1\\t=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x=1\\2^x=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\Leftrightarrow4^x-2^x+m>0;\forall x\)

Đặt \(2^x=t>0\Rightarrow t^2-t+m>0;\forall t>0\)

\(\Rightarrow m>-t^2+t\Rightarrow m>\max\limits_{t>0}\left(-t^2+t\right)=\dfrac{1}{4}\)

Vậy \(m>\dfrac{1}{4}\)

\(y'=\left(2x-2\right)f'\left(x^2-2x\right)=0\Rightarrow\left\{{}\begin{matrix}x=1\\f'\left(x^2-2x\right)=0\end{matrix}\right.\)

Xét \(f'\left(x^2-2x\right)=0\Rightarrow\left\{{}\begin{matrix}x^2-2x=-2\\x^2-2x=1\\x^2-2x=3\end{matrix}\right.\)

\(\Rightarrow x=\left\{-1;1-\sqrt{2};1+\sqrt{2};3\right\}\)

Bảng xét dấu: 

Hàm có 2 cực tiểu (sao khác cả 2 đáp án khoanh thế kia)

\(log_{\sqrt{10}}30=\dfrac{log_230}{log_2\sqrt{10}}=\dfrac{log_22+log_23+log_25}{\dfrac{1}{2}\left(log_22+log_25\right)}=\dfrac{2\left(1+a+b\right)}{1+b}\)

a.

\(y'=\dfrac{\left(1+\sqrt{3x-1}\right)'}{1+\sqrt{3x-1}}=\dfrac{3}{2\left(1+\sqrt{3x-1}\right)\sqrt{3x-1}}\)

b.

\(y'=\dfrac{\left(2sin^2x-1\right)'}{\left(2sin^2x-1\right).ln10}=\dfrac{2sin2x}{\left(2sin^2x-1\right)ln10}\)

c.

\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)