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Intersection and merger of the first-year mathematics collections in senior high schools (see inside)
(1) x 2+x-4 = 2 gives x=2 or -3 or 2x+ 1=2 then x= 1/2.

(2) Because 1/3 is the solution set of A and B, m=-20/3 n=-5/9 is brought in, so the solution set of A is {1/3,3} and the solution set of B is {1/3,-5/6.

(3) Formulating first to obtain (x+p/2+1) 2 = (p+2) 2/4-1,

When (p+2) 2/4- 1 < 0, the set A is an empty set, and A passes through the R+= empty set, and the condition holds, -4 < p < 0,-(1).

When (p+2) 2/4- 1 ≥ 0, at this time, when p≤-4 and p≥0, A intersects with R+= an empty set, then there is (p+2) 2/4-65433 under the root sign.

And when p+2≥0, the two sides are squared and the inequality does not change direction:

Then, under the radical number (p+2) 2/4- 1 < (p+2) 2/4, and then,-1 < 0, then the solution of p is p≥-2 and p≥0. When p=0, it should be checked, because in inequality.

And when p+2 < 0

Then, if the radical number is (p+2) 2/4- 1 < (p+2)/2, then the inequality is not established and the solution of p is an empty set.

To sum up, the range of p at this time is: p ≥ 0 —— (2).

It can be seen from (1)(2) that the range of p is p >-4.