Đáp án:
$a)\underset{(-\infty;1)}{max \ } y=2 \Leftrightarrow x=0\\ c)\underset{(-\infty;0)}{min \ } y=-\dfrac{1}{10} \Leftrightarrow x=-5\\ e)\underset{\left(-\infty;\tfrac{1}{2}\right)}{max \ } y=-1 \Leftrightarrow x=0.$
Giải thích các bước giải:
$a)y=\dfrac{x^2+2x-2}{x-1}, x<1\\ y'=\dfrac{(x^2+2x-2)'(x-1)-(x^2+2x-2)(x-1)'}{(x-1)^2}\\ =\dfrac{(2x+2)(x-1)-(x^2+2x-2)}{(x-1)^2}\\ =\dfrac{x^2 - 2 x}{(x-1)^2}\\ y'=0 \Leftrightarrow x=0;x=2\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-\infty&&0&&1\\\hline y'&&+&0&-&\\\hline &&&2\\y&&\nearrow&&\searrow&\\&-\infty&&&&-\infty\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-\infty;1)}{max \ } y=2 \Leftrightarrow x=0$
$c)y=\dfrac{x}{x^2+25}, x<0\\ y'=\dfrac{x'(x^2+25)-x(x^2+25)'}{(x^2+25)^2}\\ =\dfrac{x^2+25-x.2x}{(x^2+25)^2}\\ =\dfrac{25-x^2}{(x^2+25)^2}\\ y'=0 \Leftrightarrow x=\pm 5\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-\infty&&-5&&0\\\hline y'&&-&0&+&\\\hline &0&&&&0\\y&&\searrow&&\nearrow&\\&&&-\dfrac{1}{10}\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{(-\infty;0)}{min \ } y=-\dfrac{1}{10} \Leftrightarrow x=-5$
$e)y=2x+\dfrac{1}{2x-1}, x <\dfrac{1}{2}\\ y'=2-\dfrac{2}{(2x-1)^2}\\ =\dfrac{2-2(2x-1)^2}{(2x-1)^2}\\ =\dfrac{8 x - 8 x^2}{(2x-1)^2}\\ y'=0 \Leftrightarrow x=0;x=1\\ BBT:$
\begin{array}{|c|ccccccccc|} \hline x&-\infty&&0&&\dfrac{1}{2}\\\hline y'&&+&0&-&\\\hline &&&-1\\y&&\nearrow&&\searrow&\\&-\infty&&&&-\infty\\\hline\end{array}
Dựa vào $BBT \Rightarrow \underset{\left(-\infty;\tfrac{1}{2}\right)}{max \ } y=-1 \Leftrightarrow x=0.$
