xét dạng tam giác ABC thỏa:(1+cosB)/sinB=(2BC+AB)/ √(4(BC) ²-(AB) ²)

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1. Giải thích các bước giải:

   Ta có:
   $\cos B=\dfrac{BA^2+BC^2-AC^2}{2BA.BC}$
   Lại có:
   $\dfrac{1+\cos B}{\sin B}$
   $=\dfrac{1+\cos B}{\sqrt{\sin^2B}}$
   $=\dfrac{1+\cos B}{\sqrt{1-\cos^2B}}$

   $=\dfrac{1+\cos B}{\sqrt{(1-\cos B)(1+\cos B)}}$

   $=\dfrac{\sqrt{1+\cos B}}{\sqrt{1-\cos B}}$

   $=\sqrt{\dfrac{1+\cos B}{1-\cos B}}$

   $=\sqrt{\dfrac{1+\dfrac{BA^2+BC^2-AC^2}{2BA.BC}}{1-\dfrac{BA^2+BC^2-AC^2}{2BA.BC}}}$

   $=\sqrt{\dfrac{2BA.BC+BA^2+BC^2-AC^2}{2BA.BC-(BA^2+BC^2-AC^2)}}$

   $=\sqrt{\dfrac{BA^2+2BA.BC+BC^2-AC^2}{AC^2-(BA^2-2BA.BC+BC^2)}}$

   $=\sqrt{\dfrac{(BA+BC)^2-AC^2}{AC^2-(BA-BC)^2}}$

   Mặt khác:

   $\dfrac{1+\cos B}{\sin B}=\dfrac{2BC+AB}{\sqrt{4BC^2-AB^2}}$

   $\to \sqrt{\dfrac{(BA+BC)^2-AC^2}{AC^2-(BA-BC)^2}}=\dfrac{2BC+AB}{\sqrt{4BC^2-AB^2}}$

   $\to \dfrac{(BA+BC)^2-AC^2}{AC^2-(BA-BC)^2}=\dfrac{(2BC+AB)^2}{4BC^2-AB^2}$

   $\to \dfrac{(BA+BC)^2-AC^2}{AC^2-(BA-BC)^2}=\dfrac{(2BC+AB)^2}{(2BC+AB)(2BC-AB)}$

   $\to \dfrac{(BA+BC)^2-AC^2}{AC^2-(BA-BC)^2}=\dfrac{2BC+AB}{2BC-AB}$

   $\to \dfrac{(c+a)^2-b^2}{b^2-(c-a)^2}=\dfrac{2a+c}{2a-c}$ đặt $AB=c, BC=a, CA=b, a, b, c>0$

   $\to (2a-c)((c+a)^2-b^2)=(2a+c)(b^2-(c-a)^2)$

   $\to -c^3+3a^2c+cb^2+2a^3-2ab^2=-c^3+3a^2c+cb^2-2a^3+2ab^2$

   $\to 4a^3-4b^2a=0$

   $\to 4a(a^2-b^2)=0$

   $\to a^2-b^2=0$ vì $ a>0$

   $\to a^2=b^2$

   $\to a=b$ vì $a, b >0$

   $\to \Delta ABC$ cân tại $C$

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