Question

[Nguyễn Trần Tuấn Anh]

Cho a+b+c=0 (a,b,c≠0). Rút gọn biểu thức:

a) A=\(\dfrac{a^2}{bc}\)+\(\dfrac{b^2}{ca}\)+\(\dfrac{c^2}{ab}\)

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

Answer 1

[Rồng Đom Đóm]

a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{a^3+b^3+c^3}{abc}\)

\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)

b)Ta có: a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)

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

\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)

\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)

\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)

Answer 2

[Hoàng Anh Thư]

cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc

a^3+b^3+c^3=3abc

=>a^3+b^3+c^3-3abc=0

=>(a+b)^3-3ab(a+b)+c^3-3abc=0

=>[(a+b)^3+c^3]-3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0

vì a+b+c=0 nên a^3+b^3+c^3=3abc

thay kết quả vừa chúng minh vào đề bài ta đc

\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

chúc bạn học tốt ^ ^