题目内容
2.已知方程2xa+b-xa-b-ab=0是关于x的一元二次方程,求a、b的值.分析 本题根据一元二次方程的定义求解.分5种情况分别求解即可.
解答 解:∵2xa+b-xa-b+ab=0是关于x的一元二次方程,
∴①$\left\{\begin{array}{l}{a+b=2}\\{a-b=1}\end{array}\right.$,解得$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=\frac{1}{2}}\end{array}\right.$;
②$\left\{\begin{array}{l}{a+b=2}\\{a-b=0}\end{array}\right.$,解得$\left\{\begin{array}{l}{a=1}\\{b=1}\end{array}\right.$;
③$\left\{\begin{array}{l}{a+b=1}\\{a-b=2}\end{array}\right.$,解得$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=-\frac{1}{2}}\end{array}\right.$;
④$\left\{\begin{array}{l}{a+b=0}\\{a-b=2}\end{array}\right.$,解得$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$;
⑤$\left\{\begin{array}{l}{a+b=2}\\{a-b=2}\end{array}\right.$,解得$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$;
综上所述:a、b的值为:$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=\frac{1}{2}}\end{array}\right.$;$\left\{\begin{array}{l}{a=1}\\{b=1}\end{array}\right.$;$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=-\frac{1}{2}}\end{array}\right.$;$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$;$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$.
点评 本题主要考查了一元二次方程的概念.解题的关键是分5种情况讨论x的指数.
| A. | m=-1 | B. | m=0 | C. | m=1 | D. | m=2 |
| A. | x=0 | B. | x≠0 | C. | x=$\frac{2}{3}$ | D. | x$≠\frac{2}{3}$ |