ta có:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
=>\(\dfrac{a+b}{ab}+\dfrac{a+b}{ac+bc+c^2}=0\)
=>\(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)
=>\(\left(a+b\right)\left[\dfrac{ac+bc+c^2+ab}{ab\left(ac+bc+c^2\right)}\right]=0\)
=>\(\left(a+b\right).\dfrac{c\left(a+c\right)+b\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
=>\(\left(a+b\right).\dfrac{\left(c+b\right)\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
=>(a+b)(a+c)(c+b)=0
=>a+b=0<=>a=-b
a+c=0<=>a=-c
c+b=0<=>c=-b
thay a=-b ta đc \(\dfrac{1}{\left(-b\right)^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{c^3}\left(1\right)\)
\(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\left(2\right)\)
=>(1)=(2),các trường hợp còn lài cm tương tự
vậy.............
Lời giải:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\Leftrightarrow \frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
Khi đó:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3+b^3+c^3}\)\(=\frac{a^3+b^3}{a^3b^3}+\frac{a^3+b^3}{c^3(a^3+b^3+c^3)}\)
\(=(a^3+b^3)\left(\frac{1}{a^3b^3}+\frac{1}{c^3(a^3+b^3+c^3)}\right)=(a^3+b^3).\frac{a^3b^3+c^3(a^3+b^3+c^3)}{a^3b^3c^3(a^3+b^3+c^3)}\)
\(=\frac{(a^3+b^3)(b^3+c^3)(c^3+a^3)}{a^3b^3c^3(a^3+b^3+c^3)}=\frac{(a+b)(b+c)(c+a)(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)}{a^3b^3c^3(a^3+b^3+c^3)}\)
\(=0\) do \((a+b)(b+c)(c+a)=0\)
Suy ra \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\)
