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一巧设元解题就是根据问题的需要.设元得一元二次方程,又根据方程有实根的条件,使原问题得解.例1 若实数 a_1,a_2,a_3都不为零,且满足(a_1~2+a_2~2)a_4~2-2a_2(a_1+a_3)a_4+a_2~2+a_3~2=0 (1)求证:a_1、a_2、a_3为等比数列且公比为 a_4.分析:在(1)式中,设 x=a_4,则 a_4是(1)的根.由判别式得出 a_1,a_2,a_3的关系式,又由求根
One-fitting element solving problem is based on the needs of the problem. Set the element to obtain a quadratic equation, and according to the conditions of the equation have real roots, to solve the original problem. Example 1 If the real number a_1, a_2, a_3 are not zero, and Satisfaction (a_1~2+a_2~2)a_4~2-2a_2(a_1+a_3)a_4+a_2~2+a_3~2=0 (1) Prove that: a_1, a_2, and a_3 are geometric sequences and the common ratio is a_4 Analysis: In equation (1), let x = a_4, then a_4 is the root of (1). From the discriminant, we get the relation of a_1, a_2, a_3, and then find the root