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用平方法解二次根式方程,可能会产生增根,这是大家熟知的。但用立方法解三次根式方程,是否也会产生增根?这却非尽人皆晓。考察下面的例1 解方程(2χ-1)~(1/3)+(3χ-2)~(1/3)=(5χ-3)~(1/3)。解方程两边同时立方,得(5χ-3)+3·(2χ-1)~(1/3)(3χ-2)~(1/3)((2χ-1)~(1/3)+(3χ-2)~(1/3))=5χ-3把原方程代入得 ((2χ-1)(3χ-2)(5χ-3))~(1/3)=0两边再同时立方得 (2χ-1)(3χ-2)(5χ-3)=0 ∴χ_1=(1/2),χ~2=2/3,χ_3=3/5。经检验知其皆为原方程的根。例2 解方程(χ+1)~(1/3)+(2χ+5)~(1/3)=3~(1/3)。解原方程记为(1),两边同时立方,得
It is well known that solving the quadratic root equation with the flat method may generate rooting. However, if we use the standing method to solve the three root equations, will it also produce rooting? Consider the following example 1. Solve the equation (2χ-1)~(1/3)+(3χ-2)~(1/3)=(5χ-3)~(1/3). Solve both sides of the equation at the same time, and get (5χ-3)+3·(2χ-1)~(1/3)(3χ-2)~(1/3)((2χ-1)~(1/3)+ (3χ-2)~(1/3))=5χ-3 Substitution of the original equation ((2χ-1)(3χ-2)(5χ-3))~(1/3)=0 Both sides are then simultaneously cubic The results are (2χ-1)(3χ-2)(5χ-3)=0 ∴χ_1=(1/2), χ~2=2/3, and χ_3=3/5. It has been verified that all of them are the roots of the original equation. Example 2 Solution equation (χ+1)~(1/3)+(2χ+5)~(1/3)=3~(1/3). Solve the original equation as (1), both sides are cubic, get