Question

Biết : \(ab+bc+ca=0\)

Tính \(S=\frac{ab}{c^2+2ab}+\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}\)

Answer 1

sửa đề thành \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)nhé

Answer 2

Ta dễ có:

\(c^2+2ab=c^2+ab+ab=c^2+ab-bc-ca=\left(c-a\right)\left(c-b\right)\)

Một cách tương tự:

\(a^2+2bc=\left(a-b\right)\left(a-c\right);b^2+2ca=\left(b-c\right)\left(b-a\right)\)

Khi đó:

\(S=\frac{ab}{\left(c-a\right)\left(c-b\right)}+\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-c\right)\left(b-a\right)}\)

Cách đơn giản nhất là quy đồng :)

Answer 3

quy đồng không ra a kid ạ

Answer 4

Vì \(ab+bc+ca=0\Rightarrow2ab=ab-bc-ca\)

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

Tương tự: \(\frac{bc}{a^2+2bc}=\frac{bc}{\left(a-b\right)\left(a-c\right)}\); \(\frac{ca}{b^2+2ca}=\frac{ca}{\left(b-a\right)\left(b-c\right)}\)

Cộng theo vế ba đẳng thức trên, ta được: \(S=\frac{ab}{c^2+2ab}+\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}\)\(=\frac{ab}{\left(c-a\right)\left(c-b\right)}+\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}\)\(=\frac{-ab}{\left(c-a\right)\left(b-c\right)}+\frac{-bc}{\left(a-b\right)\left(c-a\right)}+\frac{-ca}{\left(a-b\right)\left(b-c\right)}\)\(=-\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=-\frac{a^2b-ab^2+b^2c-bc^2+c^2a-ca^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=-\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=-\frac{\left(b-c\right)\left(a^2+bc-ab-ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=-\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

Vậy \(S=\frac{ab}{c^2+2ab}+\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}=1\)

Answer 5

\(c^2+2ab=c^2+2ab-ab-bc-ca=c^2+ab-bc-ca=\left(b-c\right)\left(a-c\right)\)

Tương tự với các biểu thức còn lại, ta có:

\(S=\frac{ab}{\left(a-c\right)\left(b-c\right)}+\frac{bc}{\left(b-a\right)\left(c-a\right)}+\frac{ca}{\left(c-b\right)\left(a-b\right)}\)

\(S=\frac{ab\left(b-a\right)+bc\left(c-b\right)+ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

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

Answer 6

Tuổi trẻ tài cao, mà a hỏi sao nick a bị trừ 5k điểm nhỉ ? :(