\(2cos^2x-4sinxcosx=0\) 

^{\(\left\{{}\begin{matrix}cosx=0\\cosx-2sinx=0\end{matrix}\right.\)\(\left\{{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\cos\left(\alpha+x\right)=0vớicos\alpha=\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)}

gọi số dó là \(\overline{abcd}\)

d=0 , luôn có 7 nên cần chọn thêm 2 chữ số nữa là \(^{C^2_4}\)

suy ra có \(C^2_4.3!\) (số)

d=4 có \(C^2_4.\left(3!-2!\right)\)

A, vt= \(\overrightarrow{CD}+\overrightarrow{FA}+\overrightarrow{AB}+\overrightarrow{DE}+\overrightarrow{BC}+\overrightarrow{EF}\)

\(=\left(\overrightarrow{BC}+\overrightarrow{CD}\right)+\left(\overrightarrow{FA}+\overrightarrow{AB}\right)+\left(\overrightarrow{DE}+\overrightarrow{EF}\right)\)

\(=\overrightarrow{BD}+\overrightarrow{FB}+\overrightarrow{DF}=\overrightarrow{FD}+\overrightarrow{DF}=\overrightarrow{FF}=\overrightarrow{0}\)

B, \(vt=\overrightarrow{AC}+\overrightarrow{CD}-\left(\overrightarrow{FB}+\overrightarrow{BC}\right)-\left(\overrightarrow{EA}+\overrightarrow{AB}\right)\)\(=\overrightarrow{CD}-\overrightarrow{FB-}\overrightarrow{EA}+\overrightarrow{AC}-\overrightarrow{BC}-\overrightarrow{AB}\)\(=\overrightarrow{CD}-\overrightarrow{FB-}\overrightarrow{EA}+\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BA}\)

VÌ \(\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BA}=\overrightarrow{0}\) suy ra đpcm

6, \(^{\overrightarrow{NB}+2^{ }\overrightarrow{NA}=\overrightarrow{0}}\)nên N nằm giữa A , B và AN=1/3 AB

\(\overrightarrow{MB}-\overrightarrow{3MC}=\overrightarrow{0}\) nên M thuộc phần kéo dài của CB và BC=2 CM

chọn \(\overrightarrow{AB,}\overrightarrow{AC}\) làm vecto gốc

\(\overrightarrow{NI}=\overrightarrow{NA}+\overrightarrow{AI}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\)

\(\overrightarrow{NM}=\overrightarrow{NB}+\overrightarrow{BM}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{3}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\dfrac{5}{6}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}=\dfrac{5}{2}\left(-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{3}{5}\overrightarrow{AC}\right)\)nên \(\overrightarrow{NM}=\dfrac{5}{2}\overrightarrow{NI}\) suy ra N, M, I thẳng hàng

E={1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15}

A={3; 6; 9; 12; 15}

B={ 1; 2; 3; 4; 6; 12}

A\(\cup\)B={1; 2; 3; 4; 6; 9; 12; 15}

A\(\cap\)B={3; 6; 12}

A/B={9; 15}

B/A={1; 2; 4}

(A/B)\(\cup\)(B/A)={1; 2; 4; 9; 15}

bài 2: \(\overrightarrow{NC}+\overrightarrow{MC}=\overrightarrow{NC}+\overrightarrow{AN}=\overrightarrow{AC}\)

\(\overrightarrow{AM}+\overrightarrow{CD}=\overrightarrow{NC}+\overrightarrow{CD}=\overrightarrow{ND}\)

\(\overrightarrow{AD}+\overrightarrow{NC}=\overrightarrow{AD}+\overrightarrow{AM}=\overrightarrow{AE}\) ( lấy thêm E sao cho C là trung điểm ME)

b, AMCN là hbh nên \(\overrightarrow{AM}+\overrightarrow{AN}=\overrightarrow{AC}\)

ABCD là hbh nên \(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\)

Suy ra đpcm