用待定系数法因为f(x)为一次函数,设f(x)=kx+b∴f(x+1)=k(x+1)+b=kx+k+b（解析式中x用x+1替换）∴f(x-1)=k(x-1)+b=kx-k+b（解析式中x用x-1替换）∵3f(x+1)-2f(x-1)=2x+17∴3kx+3k+3b-2kx+2k-2b=2x+17∴kx+（5k+b）=2x...