Question

[Henry.]

Từ câu 01 đến câu 02.

Answer 1

[Nguyễn Việt Lâm]

1.a

\(\left|-x^2+x-4\right|>4\Leftrightarrow\left[{}\begin{matrix}-x^2+x-4>4\\-x^2+x-4< -4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+8< 0\left(vô-nghiệm\right)\\x^2-x>0\end{matrix}\right.\)

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

b.

\(sin2A+sin2B+sin2C=2sin\left(A+B\right)cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)

\(=2sinC\left[cos\left(A-B\right)+cosC\right]=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)

\(=2sinC.\left(-2sinA.sin\left(-B\right)\right)=4sinA.sinB.sinC\)

Answer 2

[Nguyễn Việt Lâm]

2.

\(\overrightarrow{AC}=\left(5;-3\right)\Rightarrow\) đường cao BH nhận (5;-3) là 1 vtpt

Phương trình BH:

\(5\left(x-1\right)-3\left(y+1\right)=0\Leftrightarrow5x-3y-8=0\)

b.

Gọi G là trọng tâm tam giác ABC \(\Rightarrow G\left(0;2\right)\)

\(\overrightarrow{BC}=\left(1;3\right)\Rightarrow\) phương trình BC có dạng:

\(3\left(x-1\right)-1\left(y+1\right)=0\Leftrightarrow3x-y-4=0\)

\(R=d\left(G;BC\right)=\dfrac{\left|3.0-1.2-4\right|}{\sqrt{3^2+\left(-1\right)^2}}=\dfrac{6}{\sqrt{10}}\Rightarrow R^2=\dfrac{18}{5}\)

Phương trình: \(x^2+\left(y-2\right)^2=\dfrac{18}{5}\)