已知x=1÷(2√a)+√a÷2 (a>1),求证:2√(x²-1)÷[x-√(x²-1)]=a-1

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已知x=1÷(2√a)+√a÷2 (a>1),求证:2√(x²-1)÷[x-√(x²-1)]=a-1
已知x=1÷(2√a)+√a÷2 (a>1),求证:2√(x²-1)÷[x-√(x²-1)]=a-1

已知x=1÷(2√a)+√a÷2 (a>1),求证:2√(x²-1)÷[x-√(x²-1)]=a-1
x=1/2(√a + 1/√a)
=>
√(x²-1)
=√(1/4*(a+1/a-2)
=1/2|√a - 1/√a|
=1/2(√a - 1/√a)(因为a>1)
=>
2√(x²-1)÷[x-√(x²-1)]
=(2√(x²-1)*[x+√(x²-1)])÷([x-√(x²-1)][x+√(x²-1)])
=2[x√(x²-1) + x²-1]
=2[1/2(√a + 1/√a)*1/2(√a - 1/√a) + 1/4*(a+1/a-2)]
=2[1/4(a-1/a) + 1/4*(a+1/a-2)]
=a-1