Question

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

chứng minh \(a+b+c=a\times b\times c\)

Answer 1

1/a + 1/b + 1/c = 2

<=> (1/a + 1/b + 1/c) = 4

<=> 1/a^2  1/b^2 + 1/c^2  +2.(1/ab + 1/bc + 1/ca) = 4

<=> 2.(1/ab + 1/bc + 1/ca) = 4-(1/a^2  +1/b^2 + 1/c^2) = 4-2 = 2

<=> 1/ab + 1/bc + 1/ca = 1

<=> a+b+c/abc = 1

<=> a+b+c = abc = a x b x c

Tk mk nha

Answer 2

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

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

\(\Rightarrow2^2=\)\(2+2.\left(\frac{a+b+c}{abc}\right)\)

\(\Rightarrow\frac{a+b+c}{abc}=\frac{2^2-2}{2}=0\)

\(\Rightarrow a+b+c=abc\) \(\left(đpcm\right)\)

Answer 3

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

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

\(\Rightarrow2^2=2+2\left(\frac{a+b+c}{abc}\right)\)

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

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