Question

6．甲、乙两人同时解方程组$\left\{\begin{array}{l}{ax+y=3}\\{x-by=1}\end{array}\right.$，甲看错了b，求得解为$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$，乙看错了a，求得解为$\left\{\begin{array}{l}{x=-1}\\{y=3}\end{array}\right.$   试求（$\frac{a}{4}$）²⁰¹⁴+b²⁰¹⁵的值．

Answer 1

分析 把$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$代入①中求出a的值，再把$\left\{\begin{array}{l}{x=-1}\\{y=3}\end{array}\right.$ 代入②中求出b的值即可，再代入代数式解答即可．

解答 解：∵甲看错了b，求得的解为$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$，
∴把$\left\{\begin{array}{l}{x=1}\\{y=-1}\end{array}\right.$代入①得，a-1=3，解得a=4；
∵乙看错了a，求得的解为$\left\{\begin{array}{l}{x=-1}\\{y=3}\end{array}\right.$，
∴把$\left\{\begin{array}{l}{x=-1}\\{y=3}\end{array}\right.$ 代入②得-1-3b=1，解得b=-$\frac{2}{3}$，
把a=4，b=-$\frac{2}{3}$代入（$\frac{a}{4}$）²⁰¹⁴+b²⁰¹⁵=1-（$\frac{2}{3}$）²⁰¹⁵．

点评 此题考查了二元一次方程组的解，方程组的解即为能使方程组中两方程成立的未知数的值，根据这一条件求出a，b的值是本题的关键．