We have :

\(\dfrac{x}{\left(a-b\right)\left(a-c\right)}+\dfrac{x}{\left(b-a\right)\left(b-c\right)}+\dfrac{x}{\left(c-a\right)\left(c-b\right)}\)

\(=x\left(\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\right)\)

\(=x.\dfrac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

So, the equation has no roots

Chúng ta có :

x ( a - b ) ( a - c )+ x ( b - a ) ( b - c )+ x ( c - a ) ( c - b )x(a−b)(a−c)+x(b−a)(b−c)+x(c−a)(c−b)

= x ( 1 ( a - b ) ( a - c )- 1 ( a - b ) ( b - c )+ 1 ( a - c ) ( b - c ))=x(1(a−b)(a−c)−1(a−b)(b−c)+1(a−c)(b−c))

= x . b - c - a + c - a - b ( a - b ) ( a - c ) ( b - c )= 0=x.b−c−a+c+a−b(a−b)(a−c)(b−c)=0

Vì vậy, phương trình không có gốc

find x or shorten the equation ?