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方程两边同乘共轭:√(x+1) + 1
1 = [√(x+1) + 1]/[2√(1+xθ(x)) ]
=> 2√(1+xθ(x)) = [√(x+1) + 1]
两边平方
4(1+xθ(x)) = [x+2+2√(x+1)]
=> 4xθ(x) = x-2+2√(x+1)
=> θ(x) = [x-2+2√(x+1)]/(4x)
1 = [√(x+1) + 1]/[2√(1+xθ(x)) ]
=> 2√(1+xθ(x)) = [√(x+1) + 1]
两边平方
4(1+xθ(x)) = [x+2+2√(x+1)]
=> 4xθ(x) = x-2+2√(x+1)
=> θ(x) = [x-2+2√(x+1)]/(4x)
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