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a²-4a+3分之a²-4×(a²+3a+2分之a-3)﹣(a²-1分之a+3) ,其中,a=根号7
人气:258 ℃ 时间:2019-11-22 00:40:24
解答
a²-4a+3分之a²-4×(a²+3a+2分之a-3)﹣(a²-1分之a+3)
=a^2/(a^2-4a+3)-4(a-3)/(a^2+3a+2)-(a+3)/(a^2-1)
=a^2/(a-1)(a-3)-4(a-3)/(a+1)(a+2)-(a+3)/(a-1)(a+1)
=a^2/(a-1)(a-3)-[4(a-3)(a-1)+(a+3)(a+2)]/(a+1)(a+2)(a-1)
=a^2/(a-1)(a-3)-(5a^2-11a+18)/(a+1)(a+2)(a-1)
=[a^2(a+1)(a+2)-(5a^2-11a+18)(a-3)]/(a+1)(a+2)(a-1)(a-3)
=[a^4-2a^3+28a^2-51a+54]/(a+1)(a+2)(a-1)(a-3)
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