" (i) "(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) .

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)+(c)/(x-c)=2(c)/(x-c)x!=a,x!=b,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

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

Show that the equation A^(2)/(x-a)+B^(2)/(x-b)+C^(2)/(x-c)+...+H^(2)/(x-h)=k has no imaginary root,where A,B,C,...., Handa,b,c,........., handk in R

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

Statement-1: If a, b, c, A, B, C are real numbers such that a lt b lt c , then f(x) = (x-a)(x-b)(x-c) -A^(2)(x-a)-B^(2)(x-b)-C^(2)(x-c) has exactly one real root. Statement-2: If f(x) is a real polynomical and x_(1), x_(2) in R such that f(x_(1)) f(x_(2)) lt 0 , then f(x) has at least one real root between x_(1) and x_(2)

Statement-1: If a, b, c, A, B, C are real numbers such that a lt b lt c , then f(x) = (x-a)(x-b)(x-c) -A^(2)(x-a)-B^(2)(x-b)-C^(2)(x-c) has exactly one real root. Statement-2: If f(x) is a real polynomical and x_(1), x_(2) in R such that f(x_(1)) f(x_(2)) lt 0 , then f(x) has at least one real root between x_(1) and x_(2)