\(\dfrac{a}{SinA}=\dfrac{b}{SinB}=\dfrac{c}{SinC}\)

\(\Rightarrow\dfrac{4}{Sin105}=\dfrac{AC}{Sin45}=\dfrac{AB}{Sin30}=4\sqrt{6}-4\sqrt{2}\)

\(\Rightarrow\left\{{}\begin{matrix}AB=2\sqrt{6}-2\sqrt{2}\\AC=-4+4\sqrt{3}\end{matrix}\right.\) ( cm )

Vậy ..

Ta có : \(S_{ABC}=\dfrac{1}{2}bc.sinA=\dfrac{1}{2}acSinB=\dfrac{1}{2}abSinC\)

\(\Rightarrow bc.sinA=acSinB=abSinC\)

- Lấy abc chia cho cả 3 vế ta được ĐPCM

\(\Leftrightarrow\left[{}\begin{matrix}1+2\cos x=0\\3-\cos x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=-\dfrac{1}{2}\\\cos x=3\end{matrix}\right.\)

Mà \(-1\le\cos x\le1\)

\(\Rightarrow\cos x=-\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\pi+k2\pi\\x=\dfrac{4}{3}\pi+k2\pi\end{matrix}\right.\)

Vậy ...

\(A=\left\{n\in N^+/x< 901\right\}\)

\(\Leftrightarrow x^3-2x+4x^2-8-4x^2+12x+8-3x=0\)

\(\Leftrightarrow x^3+7x=0\)

\(\Leftrightarrow x\left(x^2+7\right)=0\)

Mà \(x^2+7\ge7>0\)

\(\Leftrightarrow x=0\)

Vậy ...

\(=x^2y\left(8x^2+12y+20xy^3\right)\)

- Số phần phần tử là 1 .