Question

15．方程组$\left\{\begin{array}{l}{x-y=b}\\{2x+y=0}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=-1}\\{y=a}\end{array}\right.$，求代数式$\frac{a-b}{a}$÷（a-$\frac{2ab-{b}^{2}}{a}$）的值．

Answer 1

分析 先根据题意得出a、b的值，再根据分式混合运算的法则把原式进行化简，把a，b的值代入进行计算即可．

解答 解：∵方程组$\left\{\begin{array}{l}{x-y=b}\\{2x+y=0}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=-1}\\{y=a}\end{array}\right.$，
∴$\left\{\begin{array}{l}-1-a=b\\-2+a=0\end{array}\right.$，解得，$\left\{\begin{array}{l}a=2\\ b=-3\end{array}\right.$，
∴原式=$\frac{a-b}{a}$÷$\frac{（a-b）^{2}}{a}$
=$\frac{a-b}{a}$•$\frac{a}{（a-b）^{2}}$
=$\frac{1}{a-b}$，
当$\left\{\begin{array}{l}a=2\\ b=-3\end{array}\right.$时，原式=$\frac{1}{2+3}$=$\frac{1}{5}$．

点评 本题考查的是分式的化简求值，熟知分式混合运算的法则是解答此题的关键．