Chúc bạn học tốt nhó!!
Ta có: S=\(\frac{1}{2}\)AB.AC.sinA
       => sinA= \(\frac{2S}{AB.AC}\)=\(\frac{2.12}{5.8}\)=\(\frac{3}{5}\)
=>\(\widehat{A}\)\(\approx\)36,87
Lại có: BC\(^2\)=AC^2+AB^2-2AC.AB.cosA=5^2+8^2-2.5.8.cos36,87\(\approx\)25
=> BC=5

Ta có: \(\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{d^3}+\frac{d}{a^3}\ge4\sqrt[4]{\frac{a}{b^3}\frac{b}{c^3}\frac{c}{d^3}\frac{d}{a^3}}=\frac{4}{\sqrt{abcd}}\)

\(\Rightarrow\left(\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{d^3}+\frac{d}{a^3}\right)\left(a+b\right)\left(c+d\right)\ge\frac{4}{\sqrt{abcd}}2\sqrt{ab}2\sqrt{cd}=16\)

Đẳng thức xảy ra <=> a=b=c=d

Theo định lý hàm cos

\(a^2=b^2+c^2-2bc.\cos A\Rightarrow\cos A=\frac{b^2+c^2-a^2}{2bc}\)

\(c^2=a^2+b^2-2ab.\cos C\Rightarrow\cos C=\frac{a^2+b^2-c^2}{2ab}\)

\(b^2=a^2+c^2-2ac.\cos B\Rightarrow\cos B=\frac{a^2+c^2-b^2}{2ac}\)

\(\Rightarrow a\left(c\cos C-b\cos B\right)=a\left(c.\frac{a^2+b^2-c^2}{2ab}-b.\frac{a^2+c^2-b^2}{2ac}\right)=\)

\(=\frac{c^2\left(a^2+b^2-c^2\right)-b^2\left(a^2+c^2-b^2\right)}{2bc}=\)

\(=\frac{a^2c^2+b^2c^2-c^4-a^2b^2-b^2c^2+b^4}{2bc}=\frac{\left(b^4-c^4\right)-a^2\left(b^2-c^2\right)}{2bc}=\)

\(=\frac{\left(b^2+c^2\right)\left(b^2-c^2\right)-a^2\left(b^2-c^2\right)}{2bc}=\frac{\left(b^2-c^2\right)\left(b^2+c^2-a^2\right)}{2bc}=\left(b^2-c^2\right)\cos A\left(dpcm\right)\)