120 câu trắc nghiệm cuối HK1 Toán 10 - Cánh Diều - Đề 1

Trường THPT Quế Sơn thành lập vào năm 1958

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\(3\text{x}+1<0\)

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

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

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

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

\(AB\)

\(\overrightarrow{AB}\)

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

\(\overrightarrow{BA}\)

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

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

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

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

15 là số nguyên tố

15 chia hết cho 7

15 chia hết cho 7

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

\(a\subset S\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\vec{j}=\left( 0;0 \right)\)

\(\vec{j}=\left( 0;1 \right)\)

\(\vec{j}=\left( 1;0 \right)\)

\(\vec{j}=\left( 1;1 \right)\)

\(\vec{a}=\left( -1;-4 \right)\)

\(\vec{a}=\left( 4;-1 \right)\)

\(\vec{a}=\left( -4;1 \right)\)

\(\vec{a}=\left( -1;4 \right)\)

\(\vec{c}=\left( -1;-4 \right)\)

\(\vec{c}=\left( 5;12 \right)\)

\(\vec{c}=\left( 2;5 \right)\)

\(\vec{c}=\left( 2;6 \right)\)

\({A}'\left( -2;-1 \right)\)

\({A}'\left( 1;2 \right)\)

\({A}'\left( -2;1 \right)\)

\({A}'\left( 2;-1 \right)\)

\(\overrightarrow{MN}=\left( 1;1 \right)\)

\(\overrightarrow{MN}=\left( 2;2 \right)\)

\(\overrightarrow{MN}=\left( 6;8 \right)\)

\(\overrightarrow{MN}=\left( -6;-8 \right)\)

\(I\left( 4;4 \right)\)

\(I\left( 2;2 \right)\)

\(I\left( 0;-10 \right)\)

\(I\left( 0;10 \right)\)

\(D\left( 4;3 \right)\)

\(D\left( 3;4 \right)\)

\(D\left( 4;4 \right)\)

\(D\left( 8;6 \right)\)

\(\vec{a}\cdot \vec{b}=1\)

\(\vec{a}\cdot \vec{b}=12\)

\(\vec{a}\cdot \vec{b}=7\)

\(\vec{a}\cdot \vec{b}=26\)

\({{0}^{\circ }}\)

\({{90}^{\circ }}\)

\({{180}^{\circ }}\)

\(\left( 0;3 \right)\)

\(\left( 0;-3 \right)\)

\(\left( -1;2 \right)\)

\(\left( 3;4 \right)\)

\({{\text{ }\!\!\Delta\!\!\text{ }}_{a}}=\left| a-\overline{a} \right|\)

\({{\text{ }\!\!\Delta\!\!\text{ }}_{a}}=\left| a+\overline{a} \right|\)

\({{\text{ }\!\!\Delta\!\!\text{ }}_{a}}=\left| a\cdot \overline{a}\right|\)

\({{\text{ }\!\!\Delta\!\!\text{ }}_{a}}=\left| \frac{\overline{a}}{a}\right|\)

\(\left[ 1229;1239 \right]\)

\(\left[ 1234;1239 \right]\)

\(\left[ 1229;1234 \right]\)

\(\left( 1229;1234 \right)\)

7 142 000

7 141 000

7 141 500

7 141 600

\({{\text{M}}_{0}}=12\)

\({{\text{M}}_{0}}=45\)

\({{\text{M}}_{0}}=46,5\)

\({{\text{M}}_{0}}=65\)

\({{\text{M}}_{\text{e}}}=16\)

\({{\text{M}}_{\text{e}}}=17\)

\({{\text{M}}_{\text{e}}}=18\)

\({{\text{M}}_{\text{e}}}=17,5\)

\(\overline{x}\approx 6,51\)

\(\overline{x}\approx 6,36\)

\(\overline{x}\approx 6,26\)

\(\overline{x}\approx 5,34\)