先化简再求值[(3a-7)^2-(a+5)^2]\(4a-24)其中a=0.02
问题描述：
先化简再求值[(3a-7)^2-(a+5)^2]\(4a-24)其中a=0.02
答
恒等于1
答
[（3a-7）—（a+5)]^2\(4a-24)=(2a-12)^2\2(2a-12)=1 a都不用算了
答
化简后得：原式=2a-1
代入数后可得：原式=-0.96
答
[(3a-7)^2-(a+5)^2]/(4a-24)
=[(3a-7)-(a+5)][(3a-7)+(a+5)]/(4a-24)
=(3a-7-a-5)(3a-7+a+5)/(4a-24)
=(2a-12)(4a-2)/(4a-24)
=4(a-6)(2a-1)/(4a-24)
=(a-6)(2a-1)/(a-6)
=2a-1
=2*0.02-1
=0.04-1
=-0.96
答
[(3a-7)^2-(a+5)^2]\(4a-24)
=[(3a-7+a+5)*(3a-7-a-5)]/(4a-24)
=(4a-2)*(2a-12)/(4a-24)
=(2a-1)(a-6)/(a-6)
=2a-1
=2×0.02-1
=0.04-1
=-0.96