Question

Cho a,b,c #0 giải phân thức:\(\frac{x}{\left(a-b\right).\left(a-c\right)}+\frac{x}{\left(b-a\right).\left(b-c\right)}+\frac{x}{\left(c-a\right).\left(c-b\right)}=2\)

Answer 1

Với \(a,b,c\ne0\) ta có :

\(\frac{x}{\left(a-b\right)\left(a-c\right)}+\frac{x}{\left(b-a\right)\left(b-c\right)}+\frac{x}{\left(c-a\right)\left(c-b\right)}=2\)

\(\Leftrightarrow x\left(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\right)=2\)

\(\Leftrightarrow x\left(\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-a}-\frac{1}{b-c}+\frac{1}{c-a}-\frac{1}{c-b}\right)=2\)

\(\Leftrightarrow0x=2\) ( vô lí)

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