Làm tự luận giúp m với ạ
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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.
\(\left\{{}\begin{matrix}x_{A'}=x_A+\left(-1\right)=2\\y_{A'}=y_A+3=0\end{matrix}\right.\) \(\Rightarrow A'\left(2;0\right)\)
2.
\(\overrightarrow{MP}=\left(4;2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x_{N'}=x_N+4=-4+4=0\\y_{N'}=y_N+2=1+2=3\end{matrix}\right.\)
\(\Rightarrow N'\left(0;3\right)\)
3.
\(\overrightarrow{MM'}=\left(13;7\right)\Rightarrow\overrightarrow{v}=\overrightarrow{MM'}=\left(13;7\right)\)
4.
\(\overrightarrow{MN}=\left(-2;-1\right)\Rightarrow MN=\sqrt{\left(-2\right)^2+\left(-1\right)^2}=\sqrt{5}\)
\(\Rightarrow M'N'=MN=\sqrt{5}\)
5.
Gọi G là trọng tâm ABC \(\Rightarrow G\left(2;1\right)\)
\(\overrightarrow{BC}=\left(-6;-3\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x_{G'}=2-6=-4\\y_{G'}=1-3=-2\end{matrix}\right.\) \(\Rightarrow G'\left(-4;-2\right)\)
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.
\(\overrightarrow{OA}=\left(2;2;1\right)\Rightarrow OA=\sqrt{2^2+2^2+1^2}=3\)
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 \(sin^2A=2\left(sin^2B-sin^2C\right)\)
Dạng 1
Đại số
Câu 1. B
Câu 2. B
Câu 3. D
Câu 4. B
Câu 5. B
Câu 6. C
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x=-3|y|\\ 3x+y=-3\end{matrix}\right.\)
\(\Rightarrow -9|y|+y=-3\)
Nếu $y\geq 0$ thì pt trở thành:
$-9y+y=-3$
$\Leftrightarrow y=\frac{3}{8}$
$x=-3|y|=-3.\frac{3}{8}=\frac{-9}{8}$
Nếu $y< 0$ thì pt trở thành:
$9y+y=-3\Leftrightarrow y=\frac{-3}{10}$
$x=-3|y|=-3.\frac{3}{10}=\frac{-9}{10}$
Vậy hpt có 2 nghiệm $(x,y)=(\frac{3}{8}, \frac{-9}{8}); (\frac{-3}{10}, \frac{-9}{10})$