Question

Answer 1

Ta có:

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

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

\(\Leftrightarrow b^2=c^2\)

\(\Leftrightarrow b=c\) (vì b,c >0)

Thay vào pt(2)

=> \(a^2=\dfrac{a^3-2b^3}{a-2b}\)(b=c)

\(\Leftrightarrow a^3-2a^2b=a^3-2b^3\)

\(\Leftrightarrow2b\left(a^2-b^2\right)=0\)

\(\Leftrightarrow2b\left(a-b\right)\left(a+b\right)=0\)

\(\Leftrightarrow a=b\left(vì\left\{{}\begin{matrix}b>0\\a+b>0\end{matrix}\right.\right)\)

\(\Rightarrow a=b=c\)

=> Tam giác ABC đều