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1.分式方程中的解的漏解.例1若关于x的分式方程x/(x-1)+m/(x-1)=x/(x+1)无解,求m的值.错解两边同时乘以最简公分母(x+1)(x-1)得x(x+1)+m(x+1)=x(x-1)化简得:(m+2)x=-m因为:分式方程无解所以:(x+1)(x-1)=0即:x=1或x=-1;当x=1时,m=-1当x=-1时,m的值不存在所以:m=-1时,原分式方程无解错解剖析要使分式方程无解,必须是原分式方程去分母后所得整式方程无解,或所得整式方程的解是原分式方程的增根.上述解法遗漏了所得的整式方程无
1. Leakage solutions of solutions in fractional equations Example 1 If there is no solution to the partial differential equation x / (x-1) + m / (x-1) = x / (x + 1) (X + 1) = x (x-1) by multiplying both sides of the wrong solution by the simplest common denominator (x + 1) (x- (X + 1) (x-1) = 0 ie x = 1 or x = -1; when x = 1, m = -1 When x = -1, the value of m does not exist so: m = -1, the original fractional equation without disassembly To make the fractional equation no solution, must be the original fractional equation denominator obtained after the integral equation no solution, Or the solution to the resulting integral equation is the root of the original fractional equation. The above solution omits the resulting integral equation