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