一道数学题急用啊拜托了!!
解:原式=(a+b)/(a-b)+(a-b)/(a+b)-4ab/(a^2-b^2)
=(a+b)^2/(a^2-b)^2+(a-b)^2/(a^2-b^2)-4ab/(a^2-b^2)
=[(a+b)^2+(a-b)^2-4ab]/[(a+b)(a-b)]
=(2a^2-4ab+2b^2)/ [(a+b)(a-b)]
=2(a^2-2ab+b^2)/[(a+b)(a-b)]
=2(a-b)^2/[(a+b)(a-b)]
=2(a-b)/(a+b)
∵a+5b=0
∴a=-5b
把a=-5b代进2(a-b)/(a+b)得:
2*(-5b-b)/(-5b+b)=-12b/(-4b)=3
∴当...全部
解:原式=(a+b)/(a-b)+(a-b)/(a+b)-4ab/(a^2-b^2)
=(a+b)^2/(a^2-b)^2+(a-b)^2/(a^2-b^2)-4ab/(a^2-b^2)
=[(a+b)^2+(a-b)^2-4ab]/[(a+b)(a-b)]
=(2a^2-4ab+2b^2)/ [(a+b)(a-b)]
=2(a^2-2ab+b^2)/[(a+b)(a-b)]
=2(a-b)^2/[(a+b)(a-b)]
=2(a-b)/(a+b)
∵a+5b=0
∴a=-5b
把a=-5b代进2(a-b)/(a+b)得:
2*(-5b-b)/(-5b+b)=-12b/(-4b)=3
∴当a+5b=0时,原式的值为3 。
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