Bài 1:
a)\(x-\sqrt{2x-5}\ge4\)
\(\Leftrightarrow\sqrt{2x-5}\le x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5\ge0\\x-4\ge0\\2x-5\le\left(x-4\right)^2\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge4\\x^2-10x+21\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge4\\x\in R\backslash\left(3;7\right)\end{matrix}\right.\)\(\Rightarrow x\ge7\)
Vậy..
b)\(x^2-\left(m+2\right)x+2m+4=0\) (*)
Pt có hai nghiệm pb <=>\(\Delta=m^2-4m-12>0\) \(\Leftrightarrow m\in\left(-\infty;-2\right)\cup\left(6;+\infty\right)\)
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=m+2\\x_1x_2=2m+4\end{matrix}\right.\)
Có \(x_1x_2+x_1+x_2\ge3\)
\(\Leftrightarrow2m+4+m+2\ge3\) \(\Leftrightarrow m\ge-1\)
Kết hợp với đk của m=>\(m\in\left[-1;-2\right]\cup\left(6;+\infty\right)\backslash\left\{-2\right\}\)
Bài 2:
Có \(\dfrac{\pi}{2}< \alpha< \pi\)
\(\Rightarrow cos\alpha< 0,tan\alpha< 0,cot\alpha< 0\)
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\left(\dfrac{3}{5}\right):\left(-\dfrac{4}{5}\right)=-\dfrac{3}{4}\)
\(cot\left(\alpha+2019\pi\right)=cot\alpha=\dfrac{1}{tan\alpha}=-\dfrac{4}{3}\)
Bài 3:
a) Pt đt AB \(\left\{{}\begin{matrix}quaA\left(-1;3\right)\\vtcp\overrightarrow{AB}\left(2;-2\right)\Rightarrow vtpt\overrightarrow{n}\left(2;2\right)\end{matrix}\right.\)
=> AB: \(2x+2y-4=0\Leftrightarrow x+y-2=0\)
Có \(\Delta:2x+y=0\) \(\Rightarrow\overrightarrow{n_{\Delta}}\left(2;1\right)\)
\(cos\left(\widehat{AB;\Delta}\right)=cos\left(\overrightarrow{n_{AB}};\overrightarrow{n_{\Delta}}\right)\)\(=\dfrac{\left|1.2+1.1\right|}{\sqrt{1^2+1^2}.\sqrt{2^2+1^2}}\)\(=\dfrac{3\sqrt{10}}{10}\)
b) \(M\in\Delta\Rightarrow M\left(t;-2t\right)\)
Tam giác MAB cân tại M => MA=MB
\(\Leftrightarrow MA^2=MB^2\) \(\Leftrightarrow\left(-1-t\right)^2+\left(3+2t\right)^2=\left(1-t\right)^2+\left(1+2t\right)^2\)
\(\Leftrightarrow t=-\dfrac{2}{3}\)
\(\Rightarrow M\left(-\dfrac{2}{3};\dfrac{4}{3}\right)\)
