Câu hỏi:

\(\text{cos}{{120}^{\circ }}=\frac{\sqrt{3}}{2}\)

\(\text{cos}{{120}^{\circ }}=\frac{1}{2}\)

\(\text{cos}{{120}^{\circ }}=-\frac{1}{2}\)

\(\text{cos}{{120}^{\circ }}=-\frac{\sqrt{3}}{2}\)

\(AB\)

\(\overrightarrow{AB}\)

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

\(\overrightarrow{BA}\)

\(\forall \text{x}\in \text{N}:{{\text{x}}^{2}}-2>0\)

\(\exists x\in N:{{x}^{2}}-2\ge 0\)

\(\exists \text{x}\in \text{N}:{{\text{x}}^{2}}-2>0\)

\(\forall x\in N:{{x}^{2}}-2\ge 0\)

15 là số nguyên tố

15 chia hết cho 7

15 chia hết cho 7

\(\exists x\in N:x>2\)

\(a\subset S\)

\(\left\{ a \right\}\in S\)

\(\left\{ a \right\}\in S\)

\(\left\{ a \right\}\in S\)

\(\text{M}\cap \text{N}=\left[ 0;1 \right)\)

\(\text{M}\cap \text{N}=\varnothing \)

\(\text{M}\cap \text{N}=\left( 0;1 \right]\)

\(\text{M}\cap \text{N}=\left( -2;5 \right)\)

\(\frac{1}{x}-5y\ge 1\)

\(5x-2y\ge 3\)

\(\frac{5}{x}-\frac{2}{y}<3\)

\(3x-{{y}^{2}}\ge -5\)

\(-3x+y-1\ge 0\)

\(-x+y+2\le 0\)

\(x-3y-2\ge 0\)

\(3x-y-2\le 0\)

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

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

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

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

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

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

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

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