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The solution of mathematics set in senior one.
A={ 1,2},B={ 1,a- 1}

∵B is the proper subset of A, ∴B={ 1}, which means a- 1= 1, ∴a=2.

And c is a subset of a, then C=? , or C={ 1}, or C={2}, or C={ 1, 2}

When C=? ,δ = 4-4b

When C={ 1}, it means that b= 1 meets the requirements;

When C={2}, b=0, but at this time c = {0,2}, which is not satisfactory;

When C={ 1, 2}, it obviously does not meet the requirements.

To sum up, a=2, and b≥ 1.

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