Question

Cho tam giác ABC thỏa mãn: \(a\left(a^2-b^2\right)=c\left(b^2-c^2\right)\). Tính góc B

Answer 1

Lời giải:

\(a(a^2-b^2)=c(b^2-c^2)\)

\(\Leftrightarrow a(a^2-b^2)-c(b^2-c^2)=0\)

\(\Leftrightarrow (a^3+c^3)-(ab^2+b^2c)=0\)

\(\Leftrightarrow (a+c)(a^2-ac+c^2)-b^2(c+a)=0\)

\(\Leftrightarrow (a+c)(a^2-ac+c^2-b^2)=0\)

\(\Rightarrow a^2-ac+c^2-b^2=0\) (do $a+c\neq 0$)

\(\Rightarrow b^2=a^2-ac+c^2\)

Mặt khác, theo định lý hàm cosin ta có:

\(b^2=a^2+c^2-2ac\cos B\)

\(\Rightarrow a^2-ac+c^2=a^2+c^2-2ac\cos B\)

\(\Rightarrow \cos B=\frac{1}{2}\Rightarrow \widehat{B}=60^0\)