Question

Cho \(\begin{cases}a+b+c=abc\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\end{cases}\)

CMR:

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

Answer 1

Ta có: \(a+b+c=abc\)

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

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

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

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

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

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

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

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

=>ĐPCM

Answer 2

À thấy rồi, làm nè :

Ta có 1/a^2 + 1/b^2 + 1/c^2 
= (1/a + 1/b + 1/c)^2 - 2 (1/ab + 1/ac + 1/bc) 
= 4 - 2 (c/abc + b/ abc + a/ abc) 
= 4 - 2 (a+b+c)/abc 
= 4 - 2abc / abc 
= 4 - 2 
= 2 (đpcm)