题目内容
8.已知等式|2x-3y+4|+(x+2y-5)2=0的解满足方程组$\left\{\begin{array}{l}{3bx-ay=4}\\{bx+ay=12}\end{array}\right.$,求代数式a2-2ab+b2的值.分析 根据非负性可得关于x,y的方程组,解答后代入方程组$\left\{\begin{array}{l}{3bx-ay=4}\\{bx+ay=12}\end{array}\right.$中,解得a,b的值再代入代数式即可.
解答 解:由题意可得:$\left\{\begin{array}{l}{2x-3y=-4}\\{x+2y=5}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$,
把$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{3bx-ay=4}\\{bx+ay=12}\end{array}\right.$中,
可得:$\left\{\begin{array}{l}{3b-2a=4}\\{b+2a=12}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{a=4}\\{b=4}\end{array}\right.$,
把a=4,b=4代入a2-2ab+b2=16-32+16=0.
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两方程成立的未知数的值.
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