\(A=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\left(a^3+b^3+c^3\right)\frac{1}{abc}\)

Cm với a+b+c=0 thì \(a^3+b^3+c^3=3abc\)(1) .Từ đó tính dc A, muốn cm(1) bạn xét hiệu nhé

\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(luôn đúng vì a+b+c=0)

mình cm như vầy cũng đúng phải không: \(a+b+c=0\Rightarrow a+b=-c\)

                                                            \(\)                      \(\Rightarrow\left(a+b\right)^3=-c^3\)

                                                                                       \(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

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

                                                                                        \(\Rightarrow a^3+b^3+c^3=3abc\)

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath

Học tốt=)

tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2

Ukm,mình không để ý.Sorry bn

\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

\(\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}=\dfrac{ab}{\left(a+b\right)^2-2ab-c^2}+\dfrac{bc}{\left(b+c\right)^2-2bc-a^2}+\dfrac{ca}{\left(a+c\right)^2-2ac-b^2}=\dfrac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+\dfrac{bc}{\left(a+b+c\right)\left(b+c-a\right)-2bc}+\dfrac{ac}{\left(a+b+c\right)\left(a+c-b\right)-2ac}=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}.3=-\dfrac{3}{2}\)

Ta có : \(ab+bc+ca=0\)

<=> \(abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\left(\text{vì }a;b;c\ne0\right)\)

<=> \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}.\left(-\frac{1}{c}\right)\left(\text{vì }\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\right)\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Khi đó \(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)