a.Ta có : \(x\in\left(\pi;\dfrac{3}{2}\pi\right)\Rightarrow cosx< 0\) 

\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-0,8^2}=-0,6\) 

\(tanx=\dfrac{4}{3};cotx=\dfrac{3}{4}\)

b. cos 2x = \(cos^2x-sin^2x=0,6^2-0,8^2=-0,28\)

\(P=2.cos2x=-0,56\)

\(Q=tan\left(2x+\dfrac{\pi}{3}\right)=\dfrac{tan2x+tan\dfrac{\pi}{3}}{1-tan2x.tan\dfrac{\pi}{3}}=\dfrac{tan2x+\sqrt{3}}{1-tan2x.\sqrt{3}}\)

tan 2x = \(\dfrac{2tanx}{1-tan^2x}=\dfrac{\dfrac{2.4}{3}}{1-\left(\dfrac{4}{3}\right)^2}=\dfrac{-24}{7}\) 

\(Q=\dfrac{-\dfrac{24}{7}+\sqrt{3}}{1+\dfrac{24}{7}.\sqrt{3}}\) \(=\dfrac{-24+7\sqrt{3}}{7+24\sqrt{3}}\) 

Do d // d1 : x-2y+2 = 0 \(\Rightarrow\overrightarrow{n_d}=\left(1;-2\right)\)

PTĐT d : \(1\left(x-2\right)-2\left(y-1\right)=0\Leftrightarrow x-2y=0\)

d \(\perp d_1:2x+y+1=0\Rightarrow\overrightarrow{n_d}=\overrightarrow{n_1}=\left(-1;2\right)\)

PTĐT d : \(-1\left(x+2\right)+2\left(y-1\right)=0\Leftrightarrow-x+2y-4=0\)

Đặt \(\alpha=12^o\)

Ta có : \(B=\dfrac{cos\left(\dfrac{9}{2}\pi+\alpha\right).cot\left(-3\pi+\alpha\right)}{cos\left(-5\pi+\alpha\right)}\)   \(=\dfrac{cos\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\alpha-\pi\right)}{cos\left(\alpha-\pi\right)}\)

\(=\dfrac{-sin\alpha.-cot\left(\pi-\alpha\right)}{-cos\alpha}\)  \(=\dfrac{-sin\alpha.cot\alpha}{-cos\alpha}=tan\alpha.cot\alpha=1\)

Ta có : \(y=\dfrac{3x^2+1}{x^2+3}\Rightarrow y'=\dfrac{\left(3x^2+1\right)'\left(x^2+3\right)-\left(x^2+3\right)'\left(3x^2+1\right)}{\left(x^2+3\right)^2}\)

\(\Rightarrow y'=\dfrac{6x\left(x^2+3\right)-2x\left(3x^2+1\right)}{\left(x^2+3\right)^2}=\dfrac{16x}{\left(x^2+3\right)^2}\)

Sửa lại : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{tan^22\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\left(-3\sqrt{7}\right)^2+\left(\dfrac{3\sqrt{7}}{8}\right)^2:-\dfrac{1}{8}}=\dfrac{8}{441}\)

Có : 0 < \(\alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha>0\)  \(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{4}\right)^2}=\dfrac{\sqrt{7}}{4}\)

\(\Rightarrow sin2\alpha=2sin\alpha.cos\alpha=2.\dfrac{3}{4}.\dfrac{\sqrt{7}}{4}=\dfrac{3\sqrt{7}}{8}\) 

\(cos2\alpha=1-2sin^2\alpha=1-2.\left(\dfrac{3}{4}\right)^2=-\dfrac{1}{8}\)

Ta có : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{sin2\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\dfrac{3\sqrt{7}}{8}+\dfrac{\dfrac{63}{64}}{-\dfrac{1}{8}}}=\dfrac{8}{-63+3\sqrt{7}}\)

Bài này sai đề . Lấy \(a_1=2;a_2=a_3=a_4=a_5=1\) thay vào thì : 

\(VT=2^2+1^2.4+1=9\) ; \(VP=2\left(1.4+1\right)=10\)  \(\Rightarrow VT< VP\) \(\Rightarrow\) Vô Lí 

a.\(cos\left(\dfrac{-3A+B+C}{2}\right)=cos\left(\dfrac{A+B+C}{2}-2A\right)=cos\left(90^o-2A\right)=sin2A\)

b.\(cos\left(A+B-C\right)=cos\left(A+B+C-2C\right)=cos\left(180^o-2C\right)=-cos2C\)

Ta có : \(f\left(x\right)=x\left(x+1\right)\left(x+2\right)...\left(x+2018\right)\) 

\(\Rightarrow f\left(-1004\right)=0\)

\(f'\left(-1004\right)=\)\(\lim\limits_{x\rightarrow-1004}\dfrac{f\left(x\right)-f\left(-1004\right)}{x+1004}\)  

\(=\lim\limits_{x\rightarrow-1004}\dfrac{x\left(x+1\right)\left(x+2\right)...\left(x+2018\right)}{x+1004}\)

\(=\lim\limits_{x\rightarrow-1004}x\left(x+1\right)\left(x+2\right)...\left(x+1003\right)\left(x+1005\right)...\left(x+2018\right)\)

\(=-1004.\left(-1003\right).\left(-1002\right)...\left(-1\right)1.2...1014\)

\(=1014!.1004!\)