Câu hỏi:

\(\left\{ \begin{array}{*{35}{l}} x-y<-3 \\ x.y+2y\ge -4 \\ \end{array} \right.\)

\(\left\{ \begin{array}{*{35}{l}} {{x}^{2}}-y<-3 \\ x+2y\ge -4 \\ \end{array} \right.\)

\(\left\{ \begin{array}{*{35}{l}} x-y<-3 \\ x+2x\cdot y\ge -4 \\ \end{array} \right.\)

\(\left\{ \begin{array}{*{35}{l}} {{2}^{2}}\cdot x-y<-3 \\ x+y\ge -4 \\ \end{array} \right.\)

\(\text{S}=2a\cdot b\cdot \text{sin}C\)

\(\text{S}=\frac{a.b.c}{4r}\)

\(\frac{b}{\text{sin}B}=2R\)

\(\text{S}=p\cdot R\)

3

6

4

2

\(\overrightarrow{AD}=\overrightarrow{CB}\)

\(\overrightarrow{AB}=\overrightarrow{BC}\)

\(\overrightarrow{AB}=\overrightarrow{AD}\)

\(\overrightarrow{AB}=\overrightarrow{DC}\)

Vectơ \(\vec{0}\) cùng hướng với mọi vectơ

\(\overrightarrow{AA}=0\)

Vectơ \(\vec{0}\) cùng phương với mọi vectơ

\(\left| \overrightarrow{AB}\left| = \right|\overrightarrow{BA}\right|\)

\(\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AB}\)

\(\overrightarrow{CA}+\overrightarrow{BC}=\overrightarrow{BA}\)

\(\overrightarrow{CA}-\overrightarrow{CB}=\overrightarrow{BA}\)

\(\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{BC}\)

\(\overrightarrow{BC}=3\overrightarrow{AB}\)

\(\overrightarrow{AB}=-3\overrightarrow{CA}\)

\(\overrightarrow{BC}=-2\overrightarrow{BA}\)

\(\overrightarrow{AC}=-4\overrightarrow{AB}\)

\(\overrightarrow{MA}+\overrightarrow{MB}=\overrightarrow{MI}\)

\(\overrightarrow{MA}+\overrightarrow{MB}=3\overrightarrow{MI}\)

\(\overrightarrow{MI}=\frac{1}{2}\left(\overrightarrow{MA}+\overrightarrow{MB} \right)\)

\(\overrightarrow{MA}+\overrightarrow{MB}=-2\overrightarrow{MI}\)

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{AB}\)

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec{0}\)

\(\overrightarrow{GA}+\overrightarrow{GB}=\overrightarrow{GC}\)

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{AC}\)

\(\overrightarrow{GA}=2\overrightarrow{GI}\)

\(\overrightarrow{IG}=-\frac{1}{3}\overrightarrow{IA}\)

\(\overrightarrow{GB}+\overrightarrow{GC}=2\overrightarrow{GI}\)

\(\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{GA}\)