已知(3X+2)/(X-1)平方=A/(X-1)+B/(X-1)平方,求A和B的值
问题描述:
已知(3X+2)/(X-1)平方=A/(X-1)+B/(X-1)平方,求A和B的值
答
(3X+2)/(X-1)²=A/(X-1)+B/(X-1)²(3X+2)/(X-1)²=A(x-1)/(X-1)²+B/(X-1)²(3X+2)/(X-1)²=[A(x-1)+B]/(X-1)²3x+2=A(x-1)+B3x+2=Ax-A+BA=3-A+B=2B=53x+2=Ax-A+BA=3这一步的过程。。。3x+2=A(x-1)+B(就是这一步去括号啊~·) 3x+2=Ax-A+B祝学习进步,望采纳~~额,我的意思是3x+2=Ax-A+B这一步是如何求到A=3的。3x+2=A(x-1)+B 3x+2=Ax-A+B x的系数相等 A=3 常数项相等 2=-A+B