Ta có: 
bc/a²+ac/b²+ ab/c²=abc/a³+abc/b³+abc/c³ 
=abc(1/a³ + 1/b³ + 1/c³) 
=abc[(1/a + 1/b + 1/c)(1/a² + 1/b²+ 1/c²-1/ab-1/bc-1/ca)+3/abc](áp dụng HĐt trên) 
=abc.3/(abc)=3 

tk nha bạn

thank you bạn

(^_^)

\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\Leftrightarrow\dfrac{1}{c}=\dfrac{a+b}{2ab}\)

\(\Leftrightarrow2ab=c\left(a+b\right)\)

\(\Leftrightarrow ab+ab=ca+cb\)

\(\Leftrightarrow ab-cb=ca-ab\)

\(\Leftrightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{bc}{b+c}\Rightarrow\frac{abc}{c\left(a+b\right)}=\frac{abc}{b\left(a+c\right)}=\frac{abc}{a\left(b+c\right)}\)

\(\Rightarrow c\left(a+b\right)=b\left(a+c\right)\Leftrightarrow ac+bc=ab+bc\Rightarrow ac=ab\Rightarrow c=b\) (1)

\(\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Leftrightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow b=a\) (2)

\(\Rightarrow c\left(a+b\right)=a\left(b+c\right)\Leftrightarrow ac+bc=ab+ac\Rightarrow bc=ab\Rightarrow c=a\) (3)

Từ (1) ; (2) ; (3) => \(a=b=c\) (ĐPCM)

thay \(a^2=b.c\)vào biểu thức, ta có:

\(\frac{b.c+c^2}{b^2+b.c}=\frac{c.\left(c+b\right)}{b.\left(b+c\right)}=\frac{c}{b}\)