问题标题：
√{(x+1)*[(16x^2)/(x-3)^2-4(4x+3)/(x-3)]}=8化简为：9(k^2+1)^2=16(k^2-3)^2.
问题描述：
√{(x+1)*[(16x^2)/(x-3)^2-4(4x+3)/(x-3)]}=8
化简为：9(k^2+1)^2=16(k^2-3)^2.
时石回答：
√{(x+1)*[(16x^2)/(x-3)^2-4(4x+3)/(x-3)]}=8,先左右平方.
(x+1)*[(16x^2)/(x-3)^2-4(4x+3)/(x-3)]=64,左右同时除以4
(x+1)*[(4x^2)/(x-3)^2-(4x+3)/(x-3)]=16,左边同分
(x+1)*[4x^2-(4x+3)(x-3)]/(x-3)^2=16,.
(x+1)*[4x^2-4x^2+12x-3x+9]/(x-3)^2=16,.
(x+1)*(9x+9)/(x-3)^2=16,.
9(x+1)^2/(x-3)^2=16,.
9(x+1)^2=16(x-3)^2,令x=k^2得出
9(k^2+1)^2=16(k^2-3)^2
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