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题:解方程组解:观察方程组的特征易看出左边相加有1+(1+y)(1+z),且右边相加为1,故有如下简捷解法: ①+②,整理得:(1+y)(1+z)=0, ∴1+y=0,1+z=0,即y=-1,z=-1 故原主程组的解为{y=-1,z=-1。} 由上述方程组及其解,我们有一个意外的收获——韦达定理之逆定理的一个反例: 原主程组实际为:{yz=4 y+z=-5} 由韦达定理逆定理知满足此方程组即满足原方程组的y、z之(实数)值应为方程x~2+5x+4=0的两根; 从上述原方程组的解显见y=-1,z=-1,则有x~2+5x+4=0有二重根,应有△=0;
Problem: Solving equations Solution: Observing the characteristics of the equations It is easy to see that the left side is summed by 1+(1+y)(1+z), and the right side is added by 1, so there is a simple solution: 1+2. To get: (1+y)(1+z)=0, ∴1+y=0, 1+z=0, that is, y=-1, z=-1 so the solution of the original main group is {y=-1 z=-1. } From the above equations and their solutions, we have an unexpected gain—a counterexample to the inverse theorem of Vedic’s theorem: The original principal group is actually: {yz=4 y+z=-5} The inverse theorem of Vedic theorem Knowing that this equation set satisfies the y and z (real) values of the original equation set, it should be the two equations x~2+5x+4=0; obvious from the solution of the original equation set, see y=-1, z= -1, there are two roots for x~2+5x+4=0, there should be △=0;