Có 12=cosB=a2+c2−b22ac⇒ac=a2+c2−b2⇒3ac=(a+c)2−b2=(a+b+c)(a+c−b)⇒3a+b+c=a+c−bac=1a+1c−bac(∗)Ta lại có 1ac=1a(a+b)+1c(b+c)⇔a2+c2−b2=ac(luôn đúng) Do đó $(*)\Leftrightarrow \frac{3}{a+b+c}=[\frac 1a -\frac{b}{a(a+b)}]+[\frac 1c-\frac c{b+c}]=\frac 1{a+b}+\frac 1{b+c}$ (đpcm)
Có
12=cosB=a2+c2−b22ac⇒ac=a2+c2−b2⇒3ac=(a+c)2−b2=(a+b+c)(a+c−b)⇒3a+b+c=a+c−bac=1a+1c−bac(∗)Ta lại có
1ac=1a(a+b)+1c(b+c)⇔a2+c2−b2=ac(luôn đúng) Do đó $(*)\Leftrightarrow \frac{3}{a+b+c}=[\frac 1a -\frac{b}{a(a+b)}]+[\frac 1c-\frac
b{c
(b+c
)}]=\frac 1{a+b}+\frac 1{b+c}$ (đpcm)