Mk sẽ làm theo đề bài mà bạn nói dưới bình luận câu trả lời của bạn @Hồng Phúc Nguyễn.

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)

a) Thay (1) vào đề:

\(VT=\dfrac{a+2006b}{a-2006b}=\dfrac{bk+2006b}{bk-2006b}=\dfrac{b\left(k+2006\right)}{b\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(VP=\dfrac{c+2006d}{c-2006d}=\dfrac{dk+2006d}{dk-2006d}=\dfrac{d\left(k+2006\right)}{d\left(k-2006\right)}=\dfrac{k+2006}{k-2006}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}.\)

b) Thay (1) vào đề:

\(VT=\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{2006\left(bk+dk\right)}{2006bk}=\dfrac{bk+dk}{bk}=\dfrac{k\left(b+d\right)}{bk}=\dfrac{b+d}{b}\)

\(VP=\dfrac{b+d}{b}\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{2006\left(a+c\right)}{2006a}=\dfrac{b+d}{b}\rightarrowđpcm\).

\(\dfrac{a+2006b}{a-2006b}=\dfrac{c+2006d}{c-2006d}\)

\(\Leftrightarrow\)(a+2006b)(c-2006d)=(c+2006d)(a-2006b)

a(c-2006d)+2006b(c-2006d)=c(a-2006b)+2006d(a-2006b)

ac-2006ad+2006bc-4024036bd=ac-2006bc+2006ad-4024036bd

(ac-2006ad+2006bc-402436bd)-(ac-2006bc+2006ad-4024036bd=0

Suy ra 2 đẳng thức trên =nhau

Đề là gì vậy bạn???????!!!!!!!!!!!!!!!!

2)

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

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

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

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-cb-ac\right)\)

\(\Rightarrow a+b+c=0\)

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

\(\Rightarrow N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(\Rightarrow N=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)

\(\Rightarrow N=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)

\(\Rightarrow N=-1\)

Bài 1:

Thay 2006 = abc vào biểu thức A ,có :

\(\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc^2}{ac+abc^2+abc}\)

\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a\left(1+b+bc\right)}+\dfrac{c.abc}{c\left(a+ab+abc\right)}\)

\(=\dfrac{a}{a+ab+abc}+\dfrac{ab}{a+ab+abc}+\dfrac{abc}{a+ab+abc}\)

\(=\dfrac{a+ab+abc}{a+ab+abc}=1\)

Vậy tại abc = 2006 giá trị biểu thức A là 1

E xin ủng hộ cách khác cho bài 2 :(

Áp dụng hđt mở rộng ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)

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

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

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=ab+bc+ac\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

\(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)

Với \(a=b=c\) ta có: \(N=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)

Với \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\) ta có: \(N=\dfrac{-abc}{abc}=-1\)

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

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

\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)

\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)

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

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

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

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

- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)

- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)

- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)

Vậy \(P=0\)

Từ \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)

=> \(\dfrac{2005a-2006b}{2005c-2006d}=\dfrac{2006c+2007d}{2006a+2007b}\) (1)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{d}{b}\)

=> \(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{2005a}{2005c}=\dfrac{2006b}{2006d}=\dfrac{2005a+2006b}{2005c+2006d}\) (2)

Từ \(\dfrac{a}{c}=\dfrac{b}{d}\)

=> \(\dfrac{2006a}{2006c}=\dfrac{2007d}{2007b}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{2006a}{2006c}=\dfrac{2007b}{2007d}=\dfrac{2006a-2007d}{2006c-2007b}\) (3)

Từ (1),(2),(3) => \(\dfrac{2005a-2006b}{2006c+2007d}=\dfrac{2005c-2006d}{2006a+2007b}\)