VD5:

\(A\cap B\ne\varnothing\Leftrightarrow2m-1\le m+3\)

\(\Leftrightarrow m\le4\)

VD6:

\(A\cap B\ne\varnothing\Leftrightarrow9a>\dfrac{4}{a}\)

\(\Leftrightarrow\dfrac{9a^2-4}{a}>0\)

\(\Leftrightarrow9a^2-4< 0\) (do \(a< 0\))

\(\Rightarrow-\dfrac{2}{3}< a< \dfrac{2}{3}\)

\(\Rightarrow-\dfrac{2}{3}< a< 0\)

VD6:

\(A\cap B=A\Leftrightarrow A\subset B\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1\ge m\\3\le m+5\end{matrix}\right.\) \(\Rightarrow-2\le m\le-1\)

1.

\(\dfrac{1-cosx+cos2x}{sin2x-sinx}=\dfrac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}\)

\(=\dfrac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\dfrac{cosx}{sinx}=cotx\)

2.

\(\dfrac{1+tan^4x}{tan^2x+cot^2x}=\dfrac{1+tan^4x}{tan^2x+\dfrac{1}{tan^2x}}=\dfrac{1+tan^4x}{\dfrac{tan^4x+1}{tan^2x}}=tan^2x\)

3.

\(sin^4x+cos^4x=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-2sin^2x.cos^2x\)

4.

Áp dụng câu 3:

\(sin^4x+cos^4x=1-2sin^2x.cos^2x\)

\(=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2\)

\(=1-\dfrac{1}{2}sin^22x\)

5.

\(sin\left(x+y\right)sin\left(x-y\right)=\dfrac{1}{2}cos\left[\left(x-y\right)-\left(x+y\right)\right]-\dfrac{1}{2}cos\left[\left(x-y\right)+\left(x+y\right)\right]\)

\(=\dfrac{1}{2}\left(cos2y-cos2x\right)=\dfrac{1}{2}\left(1-2sin^2y\right)-\dfrac{1}{2}\left(1-2sin^2x\right)\)

\(=sin^2x-sin^2y\)

6.

\(tanx+cotx=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}\)

\(=\dfrac{1}{sinx.cosx}=\dfrac{2}{2sinx.cosx}=\dfrac{2}{sin2x}\)

A

C

giải thích ?

a: A={\(x\in N\)|\(\left\{{}\begin{matrix}x< 20\\x⋮5\end{matrix}\right.\)}

b: B={\(x\in N\)|\(\left\{{}\begin{matrix}x=4k+1\\k\in N\end{matrix}\right.\)}

a.

Áp dụng định lý hàm cos:

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=\sqrt{19}\)

Áp dụng công thức trung tuyến:

\(AM=\sqrt{\dfrac{2\left(AB^2+AC^2\right)-BC^2}{4}}=\dfrac{\sqrt{7}}{2}\)

b.

\(2\overrightarrow{IA}+\overrightarrow{IB}=\overrightarrow{0}\Leftrightarrow2\overrightarrow{IB}+2\overrightarrow{BA}+\overrightarrow{IB}=\overrightarrow{0}\Leftrightarrow\overrightarrow{IB}=\dfrac{2}{3}\overrightarrow{AB}\Rightarrow\overrightarrow{BI}=-\dfrac{2}{3}\overrightarrow{AB}\)

\(\overrightarrow{JB}-2\overrightarrow{JC}=\overrightarrow{0}\Rightarrow\overrightarrow{JB}=2\overrightarrow{JC}=2\overrightarrow{JB}+2\overrightarrow{BC}\Rightarrow\overrightarrow{JB}=-2\overrightarrow{BC}\Rightarrow\overrightarrow{BJ}=2\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{BI}.\overrightarrow{BJ}=-\dfrac{2}{3}\overrightarrow{AB}.2\overrightarrow{BC}=-\dfrac{4}{3}\overrightarrow{AB}.\overrightarrow{BC}=-\dfrac{4}{3}\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\dfrac{4}{3}AB^2-\dfrac{4}{3}\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{4}{3}.2^2-\dfrac{4}{3}.2.3.cosA=\dfrac{28}{3}\)

Độ dài IJ:

Ta có: \(\overrightarrow{BI}=-\dfrac{2}{3}\overrightarrow{AB}\Rightarrow BI=\dfrac{2}{3}AB=\dfrac{4}{3}\)

\(\overrightarrow{BJ}=2\overrightarrow{BC}\Rightarrow BJ=2BC=2\sqrt{19}\)

Từ đó:

\(IJ^2=\overrightarrow{IJ}^2=\left(\overrightarrow{IB}+\overrightarrow{BJ}\right)^2=IB^2+BJ^2+2\overrightarrow{IB}.\overrightarrow{BJ}\)

\(=IB^2+BJ^2-2\overrightarrow{BI}.\overrightarrow{BJ}\)

\(=\left(\dfrac{4}{3}\right)^2+\left(2\sqrt{19}\right)^2-2.\dfrac{28}{3}=...\)