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

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

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

Từ 1/a + 1/b + 1/c = 2 bình phương hai vế ta có:      

   (1/a + 1/b + 1/c)² = 2² 

=> 1/a² + 1/b² + 1/c² + 2(1/ab + 1/bc + 1/ ca) = 4 

=> 1/a² + 1/b² + 1/c² + 2(a + b + c)/abc = 4 (Quy đồng MTC= abc) 

=> 1/a² + 1/b² + 1/c² + 2abc/abc = 4 (Vì a + b + c = abc)

 => 1/a² + 1/b² + 1/c² + 2 = 4

 => 1/a² + 1/b² + 1/c² = 2

Vậy, P= 2

Hiểu rồi! Thanks!

ta có: a+b+c = abc

\(\Rightarrow\frac{a+b+c}{abc}=1\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)

Lại có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

                     \(2^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.1\)

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

Cho a+b+c=abc và 1/a+1/b+1/c=2.

CMR: 1/a^2 +1/b^2 +1/c^2 =2

.

\(abc=a+b+c\Leftrightarrow\frac{abc}{abc}=\frac{a+b+c}{abc}\)

\(\Leftrightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=Q\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\Rightarrow P=3^2-2Q=9-2=7\)

Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)

\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)