Ta có: \(4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

\(\Rightarrow\sqrt{4x^2-4x+5}\ge2\Rightarrow P\ge2016+2=2018\)

\(\Rightarrow P_{min}=2018\) khi \(x=\dfrac{1}{2}\)

vì mình chưa chắc x dương hay âm nên mới bỏ dô trị tuyệt đối để làm chớ

đâu có đâu do 1 bình lên vẫn là 1 mà

 \(x^2>1\Rightarrow\left|x\right|^2>1^2\Rightarrow\left|x\right|>1\) (lúc này dương hết nên mới bỏ bình được)

ĐKXĐ: \(x\ge0\)

\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)

Để \(Q\in Z\Rightarrow\sqrt{x}+3\inƯ\left(6\right)\) mà \(\sqrt{x}+3\ge3\Rightarrow\sqrt{x}+3\in\left\{3;6\right\}\)

\(\Rightarrow x\in\left\{0;9\right\}\)

ĐKXĐ: \(x\ge0\)

\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{6}{\sqrt{x}+3}\le2\Rightarrow3+\dfrac{6}{\sqrt{x}+3}\le5\)

\(\Rightarrow Q_{max}=5\) khi \(x=0\)

\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\left(a\ne\dfrac{1}{2}\right)\)

\(=\dfrac{1}{2a-1}\sqrt{5\left(a^2\right)^2\left(1-2a\right)^2}=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

Xét \(a>\dfrac{1}{2}\Rightarrow1-2a< 0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(2a-1\right)=\sqrt{5}a^2\)

Xét \(a< \dfrac{1}{2}\Rightarrow1-2a>0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(1-2a\right)=-\sqrt{5}a^2\)

\(x^2-1>0\Rightarrow x^2>1\Rightarrow\left|x\right|>1\Rightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

CO cắt AB tại D

Vì \(OA=OB=R\Rightarrow\Delta OAB\) cân tại O có \(OD\bot AB\Rightarrow D\) là trung điểm AB

\(\Rightarrow\) A và B đối xứng qua OC \(\Rightarrow\left\{{}\begin{matrix}\angle OAB=\angle OBA\\\angle CAB=\angle CBA\end{matrix}\right.\)

\(\Rightarrow\angle OAB+\angle CAB=\angle OBA+\angle CBA\Rightarrow\angle CAO=\angle CBO\)

\(\Rightarrow\angle CBO=90\Rightarrow CB\) là tiếp tuyến 

a) Vì \(AB\parallel CD\) \(\Rightarrow\angle HDC=\angle DAB=\angle KBC\) \((AD\parallel BC)\)

Xét \(\Delta CBK\) và \(\Delta CDH:\) Ta có: \(\left\{{}\begin{matrix}\angle CBK=\angle CDH\\\angle CKB=\angle CHD=90\end{matrix}\right.\)

\(\Rightarrow\Delta CBK\sim\Delta CDH\left(g-g\right)\Rightarrow\dfrac{CK}{CH}=\dfrac{CB}{CD}=\dfrac{CB}{AB}\)

Ta có: \(\angle KCH=\angle BCH+\angle KCD-\angle BCD=90+90-\angle BCD\)

\(=180-\angle BCD=\angle ABC\)

Xét \(\Delta CKH\) và \(\Delta BCA:\) Ta có: \(\left\{{}\begin{matrix}\angle KCH=\angle ABC\\\dfrac{CK}{CH}=\dfrac{CB}{AB}\end{matrix}\right.\)

\(\Rightarrow\Delta CKH\sim\Delta BCA\left(c-g-c\right)\)

b) Vì \(\Delta CKH\sim\Delta BCA\left(c-g-c\right)\Rightarrow\dfrac{HK}{AC}=\dfrac{CK}{CB}=sinCBK\)

mà \(\angle CBK=\angle BAD\Rightarrow\dfrac{HK}{AC}=sinBAD\Rightarrow HK=AC.sinBAD\)

c) \(\angle ABC=120\Rightarrow\angle CBK=60\Rightarrow sinCBK=sin60\)

\(\Rightarrow\dfrac{CK}{CB}=\dfrac{\sqrt{3}}{2}\Rightarrow CK=CB.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}.10=5\sqrt{3}\)

Ta có: \(cosCBK=\dfrac{BK}{BC}\Rightarrow cos60=\dfrac{BK}{BC}\Rightarrow BK=\dfrac{1}{2}.BC=\dfrac{1}{2}.10\)

\(=5\left(cm\right)\)

Lại có: \(\left\{{}\begin{matrix}sinCDH=sin60=\dfrac{CH}{CD}\\cosCDH=cos60=\dfrac{HD}{CD}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}CH=\dfrac{\sqrt{3}}{2}.8=4\sqrt{3}\left(cm\right)\\HD=\dfrac{1}{2}.8=4\left(cm\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}AH=AD+DH=10+4=14\left(cm\right)\\AK=AB+BK=8+5=13\left(cm\right)\end{matrix}\right.\)

Ta có: \(S_{AKCH}=S_{AKC}+S_{AHC}=\dfrac{1}{2}.AH.CH+\dfrac{1}{2}.AK.KC\)

\(=\dfrac{1}{2}.14.4\sqrt{3}+\dfrac{1}{2}.13.5\sqrt{3}=\dfrac{121}{2}\sqrt{3}\)

\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)

\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)