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: \(N=\dfrac{-abc}{abc}=-1\)
bài 1
A=
\(\dfrac{a}{ab+a+2006}+\dfrac{b}{bc+b+1}+\dfrac{2006c}{ac+2006c+2006}\)
=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc.c}{ac+abc.c+abc}\)=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc.c}{c\left(a+abc+ab\right)}\)=\(\dfrac{a}{ab+a+abc}+\dfrac{ab}{abc+ab+a}+\dfrac{abc}{a+abc+ab}\)=\(\dfrac{a+ab+abc}{ab+a+abc}=1\)
Vậy A=1
\(\dfrac{a}{ab+a+2016}+\dfrac{b}{bc+b+1}+\dfrac{2006c}{ac+2006c+2006}\)
\(=\dfrac{a}{ab+a+2006}+\dfrac{ab}{2006+ab+a}+\dfrac{2006bc}{2006+2006bc+2006b}\)
\(=\dfrac{a}{ab+a+2006}+\dfrac{ab}{a+ab+2006}+\dfrac{2006}{a+ab+2006}\)
\(=\dfrac{a+ab+2006}{a+ab+2006}=1\)
Bài 2 ) Vì \(a^3+b^3+c^3=3abc\)
Nên ta dễ dàng => \(a=b=c.\) ( Cần chưng minh bình luận ở dưới )
Thay vào N, ta có :
\(N=\left(1+\dfrac{a}{a}\right).\left(1+\dfrac{a}{a}\right).\left(1+\dfrac{a}{a}\right)=2.2.2=8\)
Vậy \(N=8\).
