Đáp án: $\Delta ABC$ cân
Giải thích các bước giải:
Ta có:
$\dfrac{1+\cos B}{\sin B}=\dfrac{2a+c}{\sqrt{4a^2-c^2}}$
$\to (\dfrac{1+\cos B}{\sin B})^2=(\dfrac{2a+c}{\sqrt{4a^2-c^2}})^2$
$\to \dfrac{1+2\cos B+\cos^2B}{\sin^2B}=\dfrac{4a^2+4ac+c^2}{4a^2-c^2}$
$\to \dfrac{1+2\cos B+1-\sin^2B}{\sin^2B}=\dfrac{4a^2+4ac+c^2}{4a^2-c^2}$
$\to \dfrac{2+2\cos B-\sin^2B}{\sin^2B}=\dfrac{4a^2+4ac+c^2}{4a^2-c^2}$
$\to \dfrac{2+2\cos B}{\sin^2B}-1=\dfrac{4a^2+4ac+c^2}{4a^2-c^2}$
$\to \dfrac{2+2\cos B}{\sin^2B}=\dfrac{4a^2+4ac+c^2}{4a^2-c^2}+1$
$\to \dfrac{2+2\cos B}{\sin^2B}=\dfrac{4a^2+4ac+c^2+4a^2-c^2}{4a^2-c^2}$
$\to \dfrac{2(1+\cos B)}{1-\cos^2B}=\dfrac{8a^2+4ac}{4a^2-c^2}$
$\to \dfrac{2(1+\cos B)}{(1+\cos B)(1-\cos B)}=\dfrac{4a(2a+c)}{(2a+c)(2a-c)}$
$\to \dfrac{2}{1-\cos B}=\dfrac{4a}{2a-c}$
$\to \dfrac1{1-\cos B}=\dfrac{2a}{2a-c}$
$\to 1-\cos B=\dfrac{2a-c}{2a}$
$\to 1-\cos B=1-\dfrac{c}{2a}$
$\to \cos B=\dfrac{c}{2a}$
$\to \dfrac{a^2+c^2-b^2}{2ac}=\dfrac{c}{2a}$
$\to a^2+c^2-b^2=c^2$
$\to a^2=b^2$
$\to a=b$
$\to \Delta ABC$ cân
