C.(1),(2),(3)

a: Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(AH^2=4\cdot9=36\)

=>AH=6(cm)

BC=BH+CH

=4+9

=13(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot CB\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{4\cdot13}=2\sqrt{13}\left(cm\right)\\AC=\sqrt{9\cdot13}=3\sqrt{13}\left(cm\right)\end{matrix}\right.\)

Xét ΔABC vuông tại A có \(tanABC=\dfrac{AC}{AB}=\dfrac{3}{2}\)

nên \(\widehat{ABC}\simeq56^0\)

b: Xét tứ giác AEHF có

\(\widehat{AEH}=\widehat{AFH}=\widehat{FAE}=90^0\)

=>AEHF là hình chữ nhật

=>AH=EF

Xét ΔHAB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\)

Xét ΔHAC vuông tại H có HF là đường cao

nên \(AF\cdot AC=AH^2\)

\(AE\cdot AB+AF\cdot AC=AH^2+AH^2=2AH^2=2FE^2\)

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

Câu 7: Hình bạn tự vẽ

Giả sử hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau. Gọi I là giao điểm của hai trung tuyến

Ta có: \(IA^2+IB^2=AB^2\)\(\Leftrightarrow\left(\dfrac{2}{3}m_a\right)^2+\left(\dfrac{2}{3}m_b\right)^2=c^2\)\(\Leftrightarrow4m_a^2+4m_b^2=9c^2\Leftrightarrow2\left(b^2+c^2\right)-a^2+2\left(a^2+c^2\right)-b^2=9c^2\Leftrightarrow a^2+b^2=5c^2\)(đúng với giả thiết)

suy ra giả sử là đúng . Vậy hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau

\(y'=4x^3-4x=0\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=1\\x=1\Rightarrow y=0\\x=-1\Rightarrow y=0\end{matrix}\right.\)

\(\Rightarrow A\left(0;1\right);B\left(1;0\right);C\left(-1;0\right)\)

\(S=\dfrac{1}{2}.\left|y_A-y_B\right|.\left|x_B-x_C\right|=\dfrac{1}{2}.1.2=1\)

\(i=120^o-90^o=30^o\)

\(i=i'\Leftrightarrow i'=30^o\)

\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\cdot\dfrac{1}{\sqrt{3}}\)

\(=\sqrt{3}\cdot\dfrac{1}{\sqrt{3}}\)

=1

you to study harder

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...eating lots of food, she is still very slim

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