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

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Theo đề ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

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

=>\(2+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=4\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)

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

=> Đpcm

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

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

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

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

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

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

⇔\(abc\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\) =abc

⇔a +b +c =abc(đpcm)

Ta có:

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

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

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

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

=> \(2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\)

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

=> \(abc.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=abc\)

=> \(c+a+b=abc\) (đpcm)

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

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

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

\(\Leftrightarrow2=2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

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

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

đpcm

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

Ta có:

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

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

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

=>2*\(\frac{a+b+c}{abc}\)=2

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

=>a+b+c=abc

\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\\ =\left(ab^2-a\right)\left(c^2-1\right)+\left(a^2b-b\right)\left(c^2-1\right)+\left(a^2c-c\right)\left(b^2-1\right)\\ =ab^2c^2-ab^2-ac^2+a+a^2bc^2-a^2b-bc^2+b+a^2b^2c-a^2c-b^2c+c\\ =abc\left(ab+bc+ac\right)-\left(a^2b+ab^2+ac^2+bc^2+a^2c+b^2c\right)+\left(a+b+c\right)\\ =abc\left(ab+bc+ca\right)+\left(a+b+c\right)+3abc-\left[\left(a^2b+ab^2+abc\right)+\left(b^2c+bc^2+abc\right)+\left(a^2c+ac^2+abc\right)\right]\\ =abc\left(ab+bc+ca\right)+abc+3abc-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\right]\\ =4abc+abc\left(ab+bc+ca\right)-\left(a+b+c\right)\left(ab+bc+ca\right)\\ =4abc+abc\left(ab+bc+ca\right)-abc\left(ab+bc+ca\right)=4abc\)

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

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

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

\(\Leftrightarrow\frac{2.\left(a+b+c\right)}{abc}=2\)

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

\(\Leftrightarrow a+b+c=abc\left(dpcm\right)\)

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

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

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

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

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

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

=> a+b+c=abc (đpcm)

Từ \(\left(1\right)\) suy ra : \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

Do \(\left(2\right)\) nên \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1,\) suy ra \(\frac{a+b+c}{abc}=1\\.\)

Do đó \(a+b+c=abc\)