Question

Cho các số a,b,c khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) và a+b+c=abc

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

Answer 1

abc=a+b+c => 1 = 1/ab + 1/bc + 1/ac 

2 = 1/a+1/b+1/c => 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2/ab + 2/ac + 2/cb 

=> 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2(1/ab + 1/ac + 1/bc) = M + 2 

=> M = 4 - 2 = 2

Mk làm bài đầu thôi,sáng nay mk làm cái tt cho

Answer 2

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

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

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

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

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

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

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