Question

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

Tính A=(a+b)(b+c)(c+a) + 9

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)=>\(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)

=>abc=(bc+ac+ab)(a+b+c)=ab²+a²b+ac²+a²c+bc²+bc²+3abc

=ab(a+b)+ac(a+c)+bc(b+c)+3abc

=>ab(a+b)+ac(a+c)+bc(b+c)+2abc=0

=>ab(a+b+c-c)+ac(a+c+c-c)+bc(b+c)+2abc=0

=>(a-c)[ac+ab)]+(b+c)(ab+bc)+2ac²+2abc=0

=>(a-c)a(c+b)+(b+c)b(a+c)+2ac(b+c)=0

=>(b+c)[(a-c)a+b(a+c)+2ac]=0

=>(b+c)(a²-ac+ab+bc+2ac)=0

=>(b+c)(a²+ab+bc+ac)=0

=>(b+c)[a(a+b)+c(a+b)]=0

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

*A=(b+c)(a+c)(a+b)+9=0+9=9.

 

Answer 2

Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

⇔ ​​​\(\dfrac{ab+ac+bc}{abc}=\dfrac{1}{a+b+c}\)​

⇔ ( ab + ac + bc )( a + b + c) = abc

⇔ a²b + ab² + b²c + bc² + a²c + ac²+ 3abc = abc 
⇔ a²b + ab² + b²c + bc² + a²c + ac²+ 2abc = 0

⇔ (a+b)(b+c)(c+a) = 0
Vậy A = 0 + 9 = 9