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Câu 1:

a: \(\sqrt{9\cdot25}=3\cdot5=15\)

b: \(=3\sqrt{2}\cdot\sqrt{2}+4\sqrt{2}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{2}\)

=6+8-10

=4

1.

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{1}{2}\Rightarrow\widehat{A}=60^o\)

\(S=\dfrac{1}{2}bc.sinA=\dfrac{1}{2}.8.5.sin60^o=10\sqrt{3}\)

\(S=\dfrac{1}{2}a.h_a=\dfrac{1}{2}.7.h_a=10\sqrt{3}\Rightarrow h_a=\dfrac{20\sqrt{3}}{7}\)

\(2R=\dfrac{a}{sinA}=\dfrac{7}{\dfrac{\sqrt{3}}{2}}=\dfrac{14\sqrt{3}}{3}\Rightarrow R=\dfrac{7\sqrt{3}}{3}\)

\(S=pr=\dfrac{a+b+c}{2}.r=10r=10\sqrt{3}\Rightarrow r=\sqrt{3}\)

\(m_a^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{129}{4}\Rightarrow m_a=\dfrac{\sqrt{129}}{2}\)

6.

a, Công thức trung tuyến:

\(AM^2=c^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{2b^2+2c^2-a^2}{4}\Rightarrow a^2=2\left(b^2-c^2\right)\)

b, \(a^2=2\left(b^2-c^2\right)\Rightarrow\dfrac{2\left(b^2-c^2\right)}{a^2}=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}-\dfrac{c^2}{a^2}\right)=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}.sin^2A-\dfrac{c^2}{a^2}.sin^2A\right)=sin^2A\)

\(\Leftrightarrow2\left(sin^2B-sin^2C\right)=sin^2A\)

Hay \(sin^2A=2\left(sin^2B-sin^2C\right)\)

18.

Do D thuộc trục hoành nên tọa độ có dạng: \(D\left(a;0;0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AD}=\left(a-3;4;0\right)\\\overrightarrow{BC}=\left(4;0;-3\right)\end{matrix}\right.\)

\(AD=BC\Leftrightarrow\left(a-3\right)^2+4^2=4^2+\left(-3\right)^2\)

\(\Rightarrow\left(a-3\right)^2=9\Rightarrow\left[{}\begin{matrix}a=0\\a=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}D\left(0;0;0\right)\\D\left(6;0;0\right)\end{matrix}\right.\)

19.

\(cos\left(\overrightarrow{a};\overrightarrow{b}\right)=\dfrac{2.\left(-1\right)+1.0+0.\left(-2\right)}{\sqrt{2^2+1^2+0^2}.\sqrt{\left(-1\right)^2+0^2+\left(-2\right)^2}}=-\dfrac{2}{5}\)

20.

\(\overrightarrow{OA}=\left(2;2;1\right)\Rightarrow OA=\sqrt{2^2+2^2+1^2}=3\)

thi ?

Lỗi r, xem lại nha

Câu 1: \(\overrightarrow{IB}+3\cdot\overrightarrow{IA}=\overrightarrow{0}\)

=>\(\overrightarrow{IB}=-3\cdot\overrightarrow{IA}\)

=>I nằm giữa A và B sao cho IB=3IA

AB=AI+BI

=>AB=AI+3AI=4AI

=>Chọn hình 4

Câu 2: Xét ΔABC có

AM là đường trung tuyến

G là trọng tâm

Do đó: A,G,M thẳng hàng

=>\(\overrightarrow{AG}=\frac23\cdot\overrightarrow{AM}=\frac23\cdot\frac12\cdot\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\frac13\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

=>Loại C

\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OG}+\overrightarrow{GA}+\overrightarrow{OG}+\overrightarrow{GB}+\overrightarrow{OG}+\overrightarrow{GC}\)

\(=3\cdot\overrightarrow{OG}+\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)=3\cdot\overrightarrow{OG}+\left(\overrightarrow{GA}+2\cdot\overrightarrow{GM}\right)\)

\(=3\cdot\overrightarrow{OG}\)

=>Chọn A

Câu 3: D

1 B

2 A

3 B

4 A

5 D

6 A

7 A

8 C

9 B

10 D

11 B