4.
a, \(p=\dfrac{a+b+c}{2}=\dfrac{9}{2}\)
Ta có \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\dfrac{3\sqrt{15}}{4}\)
Định lí cos: \(cosA=\dfrac{b^2+c^2-a^2}{2bc}=-\dfrac{1}{4}\)
\(S=\dfrac{abc}{4R}=\dfrac{3\sqrt{15}}{4}\Rightarrow R=\dfrac{abc}{3\sqrt{15}}=\dfrac{8\sqrt{15}}{15}\)
b, Ta có \(\dfrac{b^3+c^3-a^3}{b+c-a}=a^2\Leftrightarrow b^3+c^3-a^3=a^2b+a^2c-a^3\)
\(\Leftrightarrow\left(b+c\right)\left(b^2+c^2-bc-a^2\right)=0\)
\(\Leftrightarrow b^2+c^2-a^2=bc\left(1\right)\)
Khi đó: \(cotB+cotC=2cotA\)
\(\Leftrightarrow\dfrac{cosB}{sinB}+\dfrac{cosC}{sinC}=2\dfrac{cosA}{sinA}\)
\(\Leftrightarrow\dfrac{cosB}{\dfrac{b}{a}.sinA}+\dfrac{cosC}{\dfrac{c}{a}.sinA}=2\dfrac{cosA}{sinA}\)
\(\Leftrightarrow\dfrac{a.cosB}{b}+\dfrac{a.cosC}{c}=2cosA\)
\(\Leftrightarrow\dfrac{a^2+c^2-b^2}{2bc}+\dfrac{a^2+b^2-c^2}{2bc}=2.\dfrac{b^2+c^2-a^2}{2bc}\)
\(\Leftrightarrow\dfrac{2a^2}{2bc}=1\)
\(\Leftrightarrow a^2=bc\)
\(\left(1\right)\Leftrightarrow b^2+c^2-2bc=0\)
\(\Leftrightarrow\left(b-c\right)^2=0\)
\(\Leftrightarrow b=c\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\Delta ABC\) đều
