Question

Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) thì a, b, c có tính chất:

A. a.b.c = 1

B. a + b + c = 1

C. (a + b)(b + c)(c + a) = 0

D. a.b.c = 1; a + b + c = 1.

Mn giải chi tiết giúp mk vs

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) (ĐKXĐ: \(a\ne0;b\ne0;c\ne0;a+b+c\ne0\))

<=> \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)

<=> \(\dfrac{b}{ab}+\dfrac{a}{ab}+\dfrac{a+b+c}{c\left(a+b+c\right)}-\dfrac{c}{c\left(a+b+c\right)}=0\)

<=> \(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

<=> \(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

<=> \(\left(a+b\right)\left[\dfrac{c\left(a+b+c\right)}{abc\left(a+b+c\right)}+\dfrac{ab}{abc\left(a+b+c\right)}\right]=0\)

<=> \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\) [vì c(a + b + c) + ab = ac + bc + c² + ab = a(b + c) + c(b + c) = (a + c)(b + c)]

<=> (a + b)(b + c)(a + c) = 0

Answer 2

câu c : vì nhân hai vế ta được :

(a+b+c)x (ab+bc+ac)=abc

abc+a\(^2\)b+\(a^2\)c + b^2c+ab^2+abc+bc^2+ac^c+abc=abc

abc+a^2b+a^2c+ b^2c+ab^2+abc+bc^2+ac^c=0

(a+c)(a+b)(b+c)=0