Đặt:

\(\dfrac{a}{c}=\dfrac{c}{b}=k\Rightarrow\left\{{}\begin{matrix}a=ck\\c=bk\\a=bk^2\end{matrix}\right.\)

\(\dfrac{a}{b}=\dfrac{bk^2}{b}=k^2\)

\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{ck^2+bk^2}{b^2+c^2}=\dfrac{k^2\left(c^2+b^2\right)}{b^2+c^2}=k^2\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a^2+c^2}{b^2+c^2}\)

\(\Rightarrowđpcm\)

Tương tự

Ta có  : \(\frac{a}{c}=\frac{c}{b}\)

\( \implies\) \(ab=c^2\)

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

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

Super Man mà lại còn phải lên đây để hỏi bài à?

Super man hỏi bài? Nghịch lý

ok

Lời giải:

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

\(=\frac{-2(a^2+b^2+c^2-bc-ab-ac)}{(a-b)(b-c)(c-a)}=\frac{-2[(a^2+bc-ab-ac)+(b^2+ac-ba-bc)+(c^2+ab-ca-cb)]}{(a-b)(b-c)(c-a)}\)

\(=\frac{-2[(a-b)(a-c)+(b-c)(b-a)+(c-a)(c-b)]}{(a-b)(b-c)(c-a)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)

Ta có : 

\(\frac{\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\)

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

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

\(=\frac{1}{\left(a-b\right)}+\frac{1}{\left(c-a\right)}\)

 Tương tự

  \(\frac{\left(c-a\right)}{\left(b-c\right)\left(b-a\right)}\)

\(=\frac{1}{\left(b-c\right)}+\frac{1}{\left(a-b\right)}\)

\(\frac{\left(a-b\right)}{\left(c-a\right)\left(c-b\right)}\)

 \(=\frac{1}{\left(c-a\right)}+\frac{1}{\left(b-c\right)}\)

Cộng theo vế các dẳng thức trên đựoc ĐPCM 

Lam tat the ma anh van hieu