Question

3．已知$\left\{\begin{array}{l}{x=-2}\\{y=31}\end{array}\right.$是二元一次方程組$\left\{\begin{array}{l}{ax+by=1}\\{bx+ay=1}\end{array}\right.$的解，則（a+b）（a-b）的值是0．

Answer 1

分析 先把$\left\{\begin{array}{l}x=-2\\ y=31\end{array}\right.$代入二元一次方程組$\left\{\begin{array}{l}{ax+by=1}\\{bx+ay=1}\end{array}\right.$，得出關(guān)于a，b的方程組，求出a-b的值，代入代數(shù)式進(jìn)行計算即可．

解答 解：∵$\left\{\begin{array}{l}x=-2\\ y=31\end{array}\right.$是二元一次方程組$\left\{\begin{array}{l}{ax+by=1}\\{bx+ay=1}\end{array}\right.$的解，
∴$\left\{\begin{array}{l}-2a+31b=1①\\-2b+31a=1②\end{array}\right.$，
①-②得a-b=0，
∴（a+b）（a-b）=0．
故答案為：0．

點評 本題考查的是二元一次方程組的解，根據(jù)題意得出a-b的值是解答此題的關(guān)鍵．