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

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

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

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

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow2+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=4\)

\(\Rightarrow\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}=2\)

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

\(\Rightarrow a+b+c=abc\) 

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

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

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

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

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

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

.

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

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

\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

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

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

Chúc bạn học tốt.

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

1/c^2 = 2

C/m : a+ b + c = abc