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Senior one math problem
1。 Because-1≤y≤2a+3

When -2 ≤ A ≤ 0, A 2 ≤ Z ≤ 4.

So there is no solution;

When 0

So1/2 ≤ A ≤ 2;

When 2

so 2 < a≤3;

So 1/2≤a≤3.

2。 There is only one solution to this equation.

Therefore, x 2-5qx+4 = (x-u) 2 = 0 must be satisfied.

So u = 2 and q = 4/5.

That is, A={x|x=2}

The complement set of A in U is {1, 3,4,5}.

3。 Because b 2-8

So b 2

-2√2 & lt; b & lt2√2。