6*(a)/(x-a)+(b)/(x-b)=(2c)/(x-c)

Solve by factorization: (a)/(x-a)+(b)/(x-b)=(2c)/(x-c)

Long-answer type questions (L.A.) (a)/(x-a)+(b)/(x-b)=(2c)/(x-c)(xnea,b,c) .

Find x in terms of a,b and c(a)/(x-a)+(b)/(x-b)+(c)/(x-c)=2(c)/(x-c)x!=a,x!=b,x!=c

Solve by factorization: a/(x-a)+b/(x-b)=(2c)/(x-c)

Find x in terms of a , b and c : (a)/(x - a) + (b)/(x - b) = (2 c)/(x - c) , x - a , b , c

(a^2/(x-a)+b^2/(x-b)+c^2/(x-c)+a+b+c)/(a/(x-a)+b/(x-b)+c/(x-c))

If |(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))| = (a - b) (b - c) (c - a) (a + b + c) , where a,b,c are all different, then the determinant |(1,1,1),((x-a)^(2),(x-b)^(2),(x-c)^(2)),((x-b)(x-c),(x-c)(x-a),(x-a)(x-b))| vanishes when a)a + b + c = 0 b) x = (1)/(3) (a + b + c) c) x = (1)/(2) (a + b + c) d) x = a + b + c

If the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are equal, then a^(2)+b^(2)+c^(2) is equal to

Let P(x)=((x-a)(x-b))/((c-a)(c-b))c^(2)+((x-b)(x-c))/((a-b)(a-c))a^(2)+((x-c)(x-a))/((b-c)(b-a))b^(2) Prove that P(x) has the property that P(y)=y^(2) for all y in R .

Prove that (x-a)^2/((a-b)(a-c))+(x-b)^2/((b-a)(b-c))+(x-c)^2/((c-a)(c-b))=1