x+4/x =-(-x+(-4/x))≤-2√(-x)(-4/x)=-4

That is, the maximum value is -4.

2.(ax+by)^2≤(a^2+b^2)(x^2+y^2)= 1×4=4

∴-2≤ax+by≤2

That is, the maximum value is 2.

3. We know that 4-x 2 ≥ 0 and x ≠0, that is, -2≤x≤2 and x ≠ 0.

( 1)-2≤x & lt； 0 √ (4-x 2)- 1 ≥ 0

The solution is -√3≤x≤√3.

∴-√3≤x<； 0

(2)0 & lt； When x≤2, √ (4-x 2)+ 1 ≥ 0 holds.

∴0<； x≤2

According to (1)(2), the solution set of the equation is-√ 3 ≤ X.

4. the inequality is X 2-2

Let f (x) = x 2-2 and g (x) =-| x-t |, respectively, as function images, so t< t < 2.

5. If the square of u=3t/t +t+ 1 changes to ut 2+(u-3) t+u = 0.

△≥0，-(u-3)/u & lt； 0 is-3 ≤ u.