Đáp án:
2)
$ \dfrac{1}{3}<m<\dfrac{29}{12}\\
3)
Q=\dfrac{-7\sqrt{15}}{8}$
Giải thích các bước giải:
2)
Để phương trình có 2 nghiệm âm phân biệt khi $: {\left\{\begin{aligned}\Delta >0\\ S<0\\ P>0 \end{aligned}\right.}$
$\Leftrightarrow {\left\{\begin{aligned}5^2-4.(3m-1) >0\\ \dfrac{-5}{1}<0\\ \dfrac{3m-1}{1}>0 \end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned} 25-12m+4 >0\\ -5<0 (dung)\\ 3m>1 \end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}-12m >-29\\ m>\dfrac{1}{3} \end{aligned}\right.}\\
\Leftrightarrow {\left\{\begin{aligned}m <\dfrac{29}{12}\\ m>\dfrac{1}{3} \end{aligned}\right.}\\
\Leftrightarrow \dfrac{1}{3}<m<\dfrac{29}{12}\\
3)
1+\cot^2x=\dfrac{1}{\sin^2x}\\
\Rightarrow \sin^2x=\dfrac{1}{1+\cot^2x}=\dfrac{1}{1+\sqrt{15}^2}=\dfrac{1}{16}\\
\Rightarrow \sin x=\pm \dfrac{1}{4}$
Do $\pi <x<\dfrac{3\pi}{2}\Rightarrow \sin<0, \cos x<0$
$\Rightarrow \sin x=\dfrac{-1}{4}\\
\cot x=\dfrac{\cos x}{\sin x}\\
\Rightarrow \cos x=\cot x.\sin x=\sqrt{15}.\dfrac{-1}{4}=\dfrac{-\sqrt{15}}{4}\\
Q=5\cos x+3\sin 2x\\
=5.\dfrac{-\sqrt{15}}{4}+3.2\sin x\cos x\\
=\dfrac{-5\sqrt{15}}{4}+6.\dfrac{-1}{4}.\dfrac{-\sqrt{15}}{4}\\
=\dfrac{-7\sqrt{15}}{8}$
