Lời giải:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$

$\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=4$

$\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$

$\Leftrightarrow 2+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})=4$

$\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1$

$\Leftrightarrow \frac{a+b+c}{abc}=1$

$\Leftrightarrow a+b+c=abc$ (đpcm)

nhầm làm lại nha ^^

(a+b+c)^2=a^2+b^2+c^2

=>a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

=>2(ab+bc+ac)=0

=>ab+bc+ac=0

=>(ab+bc+ac)/abc=0

=>ab/abc+bc/abc+ac/abc=0

=>1/c+1/a+1/b=0

=> 1/a+1/b=-1/c

=> (1/a+1/b)^3=(-1/c)^3

=> 1/a^3+1/b^3+3/ab(1/a+1/b)=-1/c^3

=> 1/a^3+1/b^3+1/c^3+3/ab.(-1/c)=0

=> 1/a^3+1/b^3+1/c^3-3/abc=0

=> 1/a^3+1/b^3+1/c^3=3/abc (đpcm)

(a+b+c)^2=a^2+b^2+c^2

a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

2(ab+bc+ac)=0

ab+bc+ac=0

(ab+bc+ac)/abc=0

ab/abc+bc/abc+ac/abc=0

1/c+1/a+1/b=0

=> 1/a+1/b=-1/c

=> (1/a+1/b)^3=(-1/c)^3

=> 1/a^3+1/b^3+3.(1/a.)(1/b).(1/a+1/b)=-1/c^3

=> 1/a^3+1/b^3+1/c^3.3ab.(-1/c)=0

=> 1/a^3+1/b^3+1/c^3=3/abc

Bài 1 rõ thế còn gì.

Vì $abc\ne0$ nên từ $a+b+c=\dfrac1a+\dfrac1b+\dfrac1c$

suy ra $abc(a+b+c)=ab+bc+ca$.

Xét hiệu hai vế cần chứng minh:

$b(a^2-bc)(1-ac)-a(1-bc)(b^2-ac)$

$=-(a-b)\left[abc(a+b+c)-(ab+bc+ca)\right]$

$=-(a-b)\cdot0$

$=0$

Suy ra $\boxed{b(a^2-bc)(1-ac)=a(1-bc)(b^2-ac)}$

Ta có $(a+b+c)^2=a^2+b^2+c^2$

$\Rightarrow 2(ab+bc+ca)=0$

$\Rightarrow ab+bc+ca=0$

Ta có $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}
=\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}$

Mà $a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2$

$=(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-abc(a+b+c))$

$=0$

Do đó $a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2$

Suy ra $\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}$

$=\dfrac{3a^2b^2c^2}{a^3b^3c^3}$

$=\boxed{\dfrac3{abc}}$