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

\(\Leftrightarrow MA^2=MB^2\) \(\Leftrightarrow\left(-1-t\right)^2+\left(3+2t\right)^2=\left(1-t\right)^2+\left(1+2t\right)^2\)

\(\Leftrightarrow t=-\dfrac{2}{3}\)

\(\Rightarrow M\left(-\dfrac{2}{3};\dfrac{4}{3}\right)\)

Đk:\(x\ge0\)

Pt \(\Leftrightarrow2\sqrt{x}+5=36+3\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow-\sqrt{x}=22\) (vô nghiệm)

Vậy phương trình vô nghiệm

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=>\(m\in\left[-1;-2\right]\cup\left(6;+\infty\right)\backslash\left\{-2\right\}\)

\(\dfrac{\sqrt{4+\sqrt{7+4\sqrt{3}}}}{\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}\)

\(=\dfrac{\sqrt{4+\sqrt{4+4\sqrt{3}+3}}}{\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2+4\sqrt{2}+1}}}}\)\(=\dfrac{\sqrt{4+\sqrt{\left(2+\sqrt{3}\right)^2}}}{\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}}\)\(=\dfrac{\sqrt{4+2+\sqrt{3}}}{\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}}\)

\(=\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}\)\(=\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{13+30\left(\sqrt{2}+1\right)}}\)\(=\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{43+30\sqrt{2}}}\)\(=\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{25+2.5.3\sqrt{2}+\left(3\sqrt{2}\right)^2}}\)

\(=\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{\left(5+3\sqrt{2}\right)^2}}\)\(=\dfrac{\sqrt{6+\sqrt{3}}}{5+3\sqrt{2}}\)\(=\dfrac{\sqrt{6+\sqrt{3}}\left(5-3\sqrt{2}\right)}{25-\left(3\sqrt{2}\right)^2}\)\(=\dfrac{\sqrt{6+\sqrt{3}}\left(5-3\sqrt{2}\right)}{7}\)

Đt (T) có tâm I(-1;2) và R=5

Gọi phương trình tiếp tuyến kẻ từ A đến đt có dạng (d): ax+by+c=0 (a²+b²>0)

Có \(A\in\left(d\right)\Rightarrow4a+5b+c=0\) \(\Leftrightarrow c=-4a-5b\)

Có \(d_{\left(I;d\right)}=R=5\)

\(\Leftrightarrow\dfrac{\left|-a+2b+c\right|}{\sqrt{a^2+b^2}}=5\)\(\Leftrightarrow\dfrac{\left|-a+2b-4a-5b\right|}{\sqrt{a^2+b^2}}=5\)\(\Leftrightarrow\left(-5a-3b\right)^2=25\left(a^2+b^2\right)\)

\(\Leftrightarrow30ab-16b^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}8b=15a\\b=0\end{matrix}\right.\)

TH1:\(8b=15a\)

Chọn \(a=8;b=15\) =>c=-107

=>(d):\(8x+15y-107=0\)

TH2: \(b=0\Rightarrow c=-4a\)

Chọn \(c=-4;a=1\) 

=> (d): \(x-4=0\)

Do B và C có vai trò như nhau, giả sử \(B\in\left(d\right):8x+15y-107=0\), \(C\in\left(d\right):x-4=0\)

\(\Rightarrow B\left(15t+4;5-8t\right)\); \(C\left(4;y_c\right)\)

Có \(\left|\overrightarrow{IB}\right|=\left|\overrightarrow{IC}\right|=R\)

\(\Leftrightarrow\sqrt{\left(15t+5\right)^2+\left(3-8t\right)^2}=\sqrt{5^2+\left(y_c-2\right)^2}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}289t^2+102t+34=25\\25+\left(y_c-2\right)^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t=-\dfrac{3}{17}\\y_c=2\end{matrix}\right.\) \(\Rightarrow B\left(\dfrac{23}{17};\dfrac{109}{17}\right)\),\(C\left(4;2\right)\)

=> AB=3; AC=3;BC=\(\dfrac{15\sqrt{34}}{17}\)

Áp dụng ct heron tính được diện tích tam giác ABC là: \(S=\dfrac{135}{34}\) (đvdt)

Độ dài bán kính đt nội tiếp tam giác ABC là: \(r=\dfrac{S}{p}\)=...

Gọi E(\(x_E;y_E\)) là tâm đt nội tiếp tam giác ABC

=> \(\overrightarrow{EA}\left(4-x_E;5-y_E\right)\)

\(\overrightarrow{EB}\left(\dfrac{23}{17}-x_E;\dfrac{109}{17}-y_E\right)\); \(\overrightarrow{EC}\left(4-x_E;2-y_E\right)\)

Có \(BC.\overrightarrow{EA}+AB.\overrightarrow{EC}+AC.\overrightarrow{EB}=\overrightarrow{0}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15\sqrt{34}}{17}.\left(4-x_E\right)+3\left(\dfrac{23}{17}-x_E\right)+3\left(4-x_E\right)=0\\\dfrac{15\sqrt{34}}{17}\left(5-y_E\right)+3\left(\dfrac{109}{17}-y_E\right)+3\left(2-y_E\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_E=\dfrac{273+60\sqrt{34}}{102+15\sqrt{34}}\\y_E=\dfrac{75\sqrt{34}+429}{102+15\sqrt{34}}\end{matrix}\right.\)

(Toi ko bt toi sai hay đề lỗi nhưng toi chỉ làm đến đây, số xấu)

a) \(\dfrac{a^2+a+1}{a^2-a+1}=\dfrac{\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Thấy tử và mẫu của phân số đều lớn hơn 0 => \(\dfrac{a^2+a+1}{a^2-a+1}>0\)

b)\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2a+1\right)+\left(c^2-2a+1\right)\ge0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (luôn đúng với mọi a,b,c)

Dấu = xra khi a=b=c=1

a)Xét \(\left(\dfrac{a+b}{2}\right)^2-\dfrac{a^2+b^2}{2}=\)\(\dfrac{a^2+2ab+b^2-2\left(a^2+b^2\right)}{4}\)\(=\dfrac{-a^2+2ab-b^2}{4}\)\(=\dfrac{-\left(a-b\right)^2}{4}\le0\forall a;b\)

\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\) (bạn ghi sai đề?) 

Dấu = xảy ra <=> a=b

b) \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)-\left(a^8+b^8\right)\left(a^4+b^4\right)\)

\(=a^{12}+a^{10}b^2+a^2b^{10}+b^{12}-\left(a^{12}+a^8b^4+a^4b^8+b^{12}\right)\)

\(=a^2b^2\left(a^8+b^8-a^6b^2-a^2b^6\right)\)

\(=a^2b^2\left(a^2-b^2\right)\left(a^6-b^6\right)=a^2b^2\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) với mọi a,b

=> \(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)\ge\left(a^8+b^8\right)\left(a^4+b^4\right)\)

Dấu = xảy ra <=>a=b

\(A=\left(\dfrac{1-a^3}{a-a^2}+1\right)\left(\dfrac{1+a^3}{1+a}-a\right):\dfrac{\left(1-a^2\right)^3}{1+a}\)(đk: \(a\ne0;a\ne1;a\ne-1\))

\(=\left[\dfrac{\left(1-a\right)\left(1+a+a^2\right)}{a\left(1-a\right)}+1\right]\left[\dfrac{\left(1+a\right)\left(1-a+a^2\right)}{1+a}-a\right]:\dfrac{\left[\left(1-a\right)\left(1+a\right)\right]^3}{1+a}\)

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

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

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

b) \(A>A^2\Leftrightarrow A\left(1-A\right)>0\) \(\Leftrightarrow0< A< 1\)

\(\Leftrightarrow0< \dfrac{1}{a\left(1-a\right)}< 1\) 

Tại  \(\dfrac{1}{a\left(1-a\right)}>0\) \(\Leftrightarrow a\left(1-a\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}a< 0\\a>1\end{matrix}\right.\)  (1)

Tại \(\dfrac{1}{a\left(1-a\right)}< 1\) \(\Leftrightarrow\dfrac{1}{a\left(1-a\right)}-1< 0\) \(\Leftrightarrow\dfrac{1-a\left(1-a\right)}{a\left(1-a\right)}< 0\)

\(\Leftrightarrow\dfrac{1-a+a^2}{a\left(1-a\right)}< 0\) \(\Leftrightarrow a\left(1-a\right)< 0\) (vì \(1-a+a^2=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall a\))

\(\Leftrightarrow\left[{}\begin{matrix}a< 0\\a>1\end{matrix}\right.\) (2)

Từ (1) (2) ;kết hợp với đk=> \(\left[{}\begin{matrix}a< 0,a\ne-1\\a>1\end{matrix}\right.\) 

\(\dfrac{x+1}{\sqrt{x+2}-1}=3\left(đk:x\ge-2;x\ne-1\right)\)

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

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(\sqrt{x+2}+1\right)}{x+1}=3\)

\(\Leftrightarrow\sqrt{x+2}+1=3\)

\(\Leftrightarrow x+2=4\) \(\Leftrightarrow x=2\) (tm)

Vậy x=2