a)Có \(b^2+c^2-a^2=cosA.2bc\)

\(S=\dfrac{1}{2}bc.sinA\)\(\Rightarrow4S=2bc.sinA\)

\(\Rightarrow\dfrac{b^2+c^2-a^2}{4S}=\dfrac{cosA.2bc}{2bc.sinA}=cotA\) (dpcm)

b) CM tương tự câu a \(\Rightarrow\dfrac{a^2+c^2-b^2}{4S}=\dfrac{cosB.2ac}{2ac.sinB}=cotB\); \(\dfrac{a^2+b^2-c^2}{4S}=\dfrac{cosC.2ab}{2ab.sinC}=cotC\)

Cộng vế với vế \(\Rightarrow cotA+cotB+cotC=\dfrac{b^2+c^2-a^2}{4S}+\dfrac{a^2+c^2-b^2}{4S}+\dfrac{a^2+b^2-c^2}{4S}\)\(=\dfrac{a^2+b^2+c^2}{4S}\) (dpcm)

c) Gọi m_a;m_b;m_c là độ dài các đường trung tuyến kẻ từ đỉnh A;B;C của tam giác ABC 

Có \(GA^2+GB^2+GC^2=\dfrac{4}{9}\left(m_a^2+m_b^2+m_b^2\right)\)\(=\dfrac{4}{9}\left[\dfrac{2\left(b^2+c^2\right)-a^2}{4}+\dfrac{2\left(a^2+c^2\right)-b^2}{4}+\dfrac{2\left(b^2+c^2\right)-a^2}{4}\right]\)

\(=\dfrac{4}{9}.\dfrac{3\left(a^2+b^2+c^2\right)}{4}=\dfrac{a^2+b^2+c^2}{3}\) (đpcm)

d) Có \(a\left(b.cosC-c.cosB\right)=ab.cosC-ac.cosB\)

\(=\dfrac{a^2+b^2-c^2}{2}-\dfrac{a^2+c^2-b^2}{2}\)

\(=b^2-c^2\) (dpcm)

a, 3 đường trung tuyến cách nhau tại trọng tâm, khoảng cách từ trọng tâm đến đỉnh bằng \(\dfrac{2}{3}\) độ dài trung tuyến đi qua đỉnh đó

Từ định lí trên ta có \(\left\{{}\begin{matrix}m_a=\dfrac{2}{3}GA\\m_b=\dfrac{2}{3}GB\\m_c=\dfrac{2}{3}GC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m_a^2=\dfrac{4}{9}GA^2\\m_b^2=\dfrac{4}{9}GB^2\\m_c^2=\dfrac{4}{9}GB^2\end{matrix}\right.\)

Đặt D = GA² + GB² + GC² 

⇒ D = m_a² + m_b² + m_c² 

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

⇒ D = \(\dfrac{a^2+b^2+c^2}{3}\)

b, cotA = \(\dfrac{cosA}{sinA}=\dfrac{\dfrac{b^2+c^2-a^2}{2bc}}{\dfrac{a}{2R}}=R.\dfrac{b^2+c^2-a^2}{abc}\)

Tương tự ta có

cotB = \(R.\dfrac{a^2+c^2-b^2}{abc}\)

cotC = \(R.\dfrac{a^2+b^2-c^2}{abc}\)

Vậy cotA + cotB + cotC = \(R.\dfrac{a^2+b^2+c^2}{abc}\) (1)

Theo công thức tính diện tích

S = \(\dfrac{abc}{4R}\) ⇒ abc = 4 . S . R

Thế vào (1) ta có

cotA + cotB + cotC = \(R.\dfrac{a^2+b^2+c^2}{4.S.R}=\dfrac{a^2+b^2+c^2}{4S}\)

a, \(\overrightarrow{GA}=-\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(\Rightarrow GA^2=\dfrac{1}{9}\left(AB^2+AC^2+2AB.AC.cosA\right)\)

\(=\dfrac{1}{9}\left(c^2+b^2+2bc.cosA\right)\)

\(=\dfrac{1}{9}\left(c^2+b^2+b^2+c^2-a^2\right)=\dfrac{2b^2+2c^2-a^2}{9}\)

Tương tự \(GB^2=\dfrac{2a^2+2c^2-b^2}{9}\); \(GC^2=\dfrac{2a^2+2b^2-c^2}{9}\)

\(\Rightarrow GA^2+GB^2+GC^2=\dfrac{a^2+b^2+c^2}{3}\)

b, \(cotA+cotB+cotC=\dfrac{cosA}{sinA}+\dfrac{cosB}{sinB}+\dfrac{cosC}{sinC}\)

\(=\dfrac{b^2+c^2-a^2}{2bcsinA}+\dfrac{a^2+c^2-b^2}{2acsinB}+\dfrac{a^2+b^2-c^2}{2absinC}\)

\(=\dfrac{b^2+c^2-a^2}{2bcsinA}+\dfrac{a^2+c^2-b^2}{2ac.\dfrac{b}{a}sinA}+\dfrac{a^2+b^2-c^2}{2ab.\dfrac{c}{a}sinA}\)

\(=\dfrac{a}{2sinA}\left(\dfrac{b^2+c^2-a^2}{abc}+\dfrac{a^2+c^2-b^2}{abc}+\dfrac{a^2+b^2-c^2}{abc}\right)\)

\(=\dfrac{a^2+b^2+c^2}{2bcsinA}=\dfrac{a^2+b^2+c^2}{4.S}\)

\(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)

\(=\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}\)

\(=\dfrac{a^2+b^2+c^2}{2abc}\) (đpcm)

a² = b² + c² - 2bc.cosA

b² = a² + c² - 2ac.cosB

c² = a² + b² - 2ab.cosC

⇒ a² + b² + c² = 2bc.cosA + 2ac.cosB + 2ab.cosC

⇒ VT =  \(\dfrac{2bc.cosA}{2abc}+\dfrac{2ab.cosC}{2abc}+\dfrac{2ac.cosB}{2abc}\)

⇒ VT = \(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\)