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![](https://rs.olm.vn/images/avt/0.png?1311)
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Gọi G là trọng tâm tam giác\(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
Đặt \(P=MA^2+MB^2+MC^2=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^2\)
\(=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)\)
\(=3MG^2+GA^2+GB^2+GC^2\)
Do \(GA^2+GB^2+GC^2\) ko đổi nên \(P_{min}\) khi \(MG_{min}\Leftrightarrow M\) là chân đường vuông góc hạ từ G xuống BC
\(\Rightarrow\dfrac{CM}{BC}=\dfrac{2}{3}\Rightarrow\dfrac{BM}{BC}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(sin\left(\text{α}-\dfrac{\Pi}{4}\right)-cos\left(\text{α}-\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.cos\dfrac{\Pi}{4}-cos\text{α}-sin\dfrac{\Pi}{4}-\left(cos\text{α}.cos\dfrac{\Pi}{4}+sin\text{α}.sin\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-sin\text{α}.\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-2\sqrt{2}}{6}\)
\(=\dfrac{-\sqrt{2}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 0 < α < π/2 nên sin α > 0, cos α > 0, tan α > 0, cot α > 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
TL:
=3828240.97059
-HT-