初二数学~~~分式方程~~~~~~~先化简再求值......
先化简再求值: (a^2+3a)/(a^2+4a+4)÷(a+3)/(a+2)-2/(a+2),其中a=(√2)-2
(a^2+3a)/(a^2+4a+4)÷(a+3)/(a+2)-2/(a+2),其中a=(√2)-2 解:(a^2+3a)/(a^2+4a+4)÷(a+3)/(a+2)-2/(a+2)= a(a+3)/(a+2)^2÷(a+3)/(a+2)-2/(a+2)= [a(a+3)/(a+2)^2]×[(a+2)/(a+3)]-2/(a+2)= a/(a+2)-2/(a+2)= (a-2)/(a+2)=(其中a=(√2)-2) (√2)-4)/√2=(2-4√2)/2=1-2√2
先化简再求值: (a^2+3a)/(a^2+4a+4)÷(a+3)/(a+2)-2/(a+2),其中a=(√2)-2 (a^2+3a)/(a^2+4a+4)÷(a+3)/(a+2)-2/(a+2) =[a(a+3)/(a+2)^2]÷[(a+3)/(a+2)]-[2/(a+2)] =[a(a+3)/(a+2)^2]*[(a+2)/(a+3)]-[2/(a+2)] =[a/(a+2)]-[2/(a+2)] =(a-2)/(a+2) =(√2-2-2)/(√2-2+2) =(√2-4)/√2 =1-2√2
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