解:枣咐
∵f(x)=x^2+4x+3
∴f(ax+b)=(ax+b)^2+4(ax+b)+3
=a^2x^2+(4a+2ab)x+b^2+4b+3
=x^2+10x+24
两个多项式携岩耐相等,那么对应系数相等
∴a^2=1,辩春4a+2ab=10
解得a=1,b=3
∴ 5a -b = 5-3 = 2
不懂再问哦
f(x)=x²+4x+3,且f(ax+b)=x²蔽弯此闹姿+10x+24,
f(ax+b)=(ax+b)²+4(ax+b)+3=a²x²+(2ab+a)x+b²+4b+3=x²+10x+24
a²宏迅=1
2ab+4a=10
b²+4b+3=24
解得:a= 1 b=3 或a= -1 b= -7
a= 1 b= 3 时: 5a-b=2
a= -1 b= -7 时:5a-b=2
则5a-b=2
(ax+b)^2+4ax+4b+3=x^2+10x+24
a^2x^2+b^2+2abx+4ax+4b-18-x^2-10x=0