Tham khảo:

Cho a≠b≠c, a+b≠c và c2+2ab-2ac-2bc=0 Hãy rút gọn \(B=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\) - Hoc24

\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)

\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)

\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)

Bài 1:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

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

\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

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

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)

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

Bài 2:

\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))

\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)

Cái đầu ko rút gọn được

Cái sau:

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

ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b 

ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)

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

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

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

M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)

M=-1-1-1=-3

Vậy với a+b+c=0 thì M=-3

\(c^2-2ac+a^2+2ab-2bc=a^2\)

\(\Rightarrow\left(a-c\right)^2+2b\left(a-c\right)=a^2\)

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

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