Let a and b belong to R, and satisfy a 2 b 2 6a 4b 12 0

Solution: 1) a 2 + b 2-6a - 4b + 12 = 0

a-3)^2+(b-2)^2=1

The parameter sinr,,b-2=cosr can be set

A 2+B 2=(sinr+3) 2+(cosr+2) 2=1+9+4+6sinr+4cosr=14+2(3sinr+2cosr)=14+2(13)sin(r+k), (k is a parameter).

It can be solved from the maximum value of the trigonometric function to obtain max=14+2(13), min=14-2(13).

2) b a=(cosr+2) (sinr+3) (using the universal formula to have only one parameter tan(r 2), which is a bit cumbersome, and there may be a better algorithm.) )

3) The A+2B solution is the same as the first question.

Answer: [5+ 2,5- 2].

ps: Triangular exchange is the basic idea of high 1, no matter how simple it is, it is not high 1. This is the simplest and a bit complicated to write.

2a+b=2

4 a+2 b=(2 2) a+2 b=2 2a+2 b>=2 root number (2 2a*2 b) = 2 root number [2 (2a+b)] = 2*2 = 4

When 2 2a = 2 b, i.e. 2a = b, take the equal sign.

2a+b=2, then hail 2a=b=1, so the equal sign can be taken.

Therefore, the source contains the minimum value of potatoes = 4

a+1/a)^2+(b+1/b)^2

a 2+1 a 2+2 +b 2+1 b 2+2(a 2+b 2) +1 a 2+1 b 2) +4= + 1 cover b) 2 +4

1/b)^2+4

Because ab<=, 1 a+1 b=(a+b) ab=1 ab >=4;

1/a +1/b)^2>=16

So (a+1 a) 2+(b+1 b) 2>=

Apparently there are (a b) 2 0, a b) 2 4ab 0·· and a 2 b 2 10, a b) split 2 2 ab 10, 4 ab 2 (a b) 2 20·· De: (a b) 2 20, Duan Qiao (a b) 2 20, Grip Yuan Meng 2 5 a b 2 5

That is, the value range of a b is 2 5, 2 5

a+b=1, then 1 a+1 b=(1 a+1 b)(a+b)=1+1+a b+b a=2+a b+b a

AB belongs to R+

Then a b+b a 2 (a b*b a)=2 is greater than or equal to 2+2=4

Take the equal sign only when a=b=1 2.

So 1 a+1 b 4 (equal sign only if a=b=1 2).

a>b 2>0, i.e. 2a>b>0, so.

t=a²＋16/[(2a－b)b]

1/4)×{2(2a－b)b＋2(2a－b)b＋16/[(2a－b)b]

2a－b)b＋16/[(2a－b)b]≥8。The condition for obtaining the equal sign is 2a b=b and (2a b)b=16 [(2a b)b], that is, 2a b=b and (2a b)b=4, i.e., 2a b=b=2, i.e., a=2, b=2 (this can be done).

So, the minimum value of a 16 [(2a b)b] is 8.

Let x=a+b

b = x-a, so a +2(x-a) = 6

3a²-4ax+2x²-6=0

a is a real number, then the equation has a solution.

Therefore, the discriminant one is equal to 0

16x²-24x²+72>=0

x²<=9

3<=x<=3

So the minimum value is 3

Cauchy inequality.

1+1 2)(a 2+2b 2) (a+b) 2 so -3 a+b 3

Therefore, the minimum value a+b=-3 is obtained when a=2b=-2.

Whether it's a multiplied by b, or a and b.

A 2 + b 2 = 4 is greater than or equal to 2 ab When a = b = root number 2 is taken as an equal sign ab max = 2 ab min = 0

The solution to this problem is as follows.

Let a=2sinx,b=2cosx x[0,2 ]ab=2sinx2cosx=2sin2x 2x[0,4] so ab[-2,2].