Question

1) \(\frac{x-b-c}{a}\) + \(\frac{x-c-a}{b}\) + \(\frac{x-a-b}{c}\)= 3

2) \(\frac{x-a}{bc}\) + \(\frac{x-b}{ac}\) + \(\frac{x-c}{ab}\)= 2 (\(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{c}\))

Answer 1

1)\(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-a-b}{c}=3\)

=>\(\frac{x-b-c}{a}-1+\frac{x-c-a}{b}-1+\frac{x-a-b}{c}-1=0\)

=>\(\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)

=>\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

Nếu x - a -b -c = 0 => phương trình có nghiệm duy nhất x = a + b + c

Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)=> Phương trình có vô số nghiệm x thuộc R