9{(x-1)的平方}=16{(x+2)的平方} 9(x-1)²-16(x+2)²=0[3(x-1)+4(x+2)][3(x-1)-4(x+2)]=0(7x+5)(-x-11)=0x=-5/7或x=-11
解：9{(x-1)²}=16{(x+2)²}开平方，得3(x-1)=±4(x+2)3(x-1)=4(x+2)或3(x-1)=-4(x+2)∴x1=-11, x2=-5/7.
