Question

[Hàn Vũ Nhi]

Chứng minh rằng nếu  \(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{c}\)=   2 và a + b + c = a.b.c. thì \(\frac{1}{a^2}\)+ \(\frac{1}{b^2}\)+ \(\frac{1}{c^2}\)= 2

Answer 1

[Nguyễn Văn Hưởng]

Ta có : a + b + c = abc

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

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

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

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

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

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

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

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

Answer 2

[Nguyệt]

\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=\frac{abc}{abc}=1\left(\text{vì }a+b+c=abc\right)\)

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

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

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

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.1=2\left(\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\right)\)

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