A and a are integers.

B and a are rational numbers.

C and a are irrational numbers.

D and a may be rational numbers or irrational numbers.

Solution: divide both sides of the equation by ab at the same time,

1b+ 2+ 1a= 2ab，

After sorting, a+bab= 2(ab- 1)ab.

So a+b= 2(ab- 1)

There is an irrational number on one side of the equation. If both a and b are rational numbers, the equation will never hold.

And because B is rational, A must be irrational.

So choose C.

2. Prove that 2.61545454 ... = 2.6154 ... cyclic decimals are rational numbers.

Proof: let x = 2.61$ \ stackrel {...} {54} $,①

Both sides are multiplied by 100,100x = 2.61$ \ stackrel { } {54} $ = 26 1.$ \ stackrel {。 } {54} $,②

②-① Germany

99x = 26 1.54-2.6 1 = 258.93，

So x=$\frac{25893}{9900}$,

Because the fraction is a rational number, X is a rational number, that is, 2.61545454 … = 2.61$ \ stackrel {..} {54} $ cyclic decimal is a rational number.

3. Three mutually unequal rational numbers can be expressed in the form of 1, a+b, a, or 0, ba, b. Try to find the value of a2000+b200 1.

Solution: ∵ Three mutually unequal rational numbers are represented by 1, a+b, a, and can also be represented by 0, ba, b,

The numbers in these two arrays are equal respectively.

∴ One of A+B and A is 0, and one of ba and B is 1, but if a=0, ba is meaningless.

∴a≠0, only a+b=0, which means a=-b, so Ba =- 1. Only b= 1, so A =- 1.

∴ Original formula = (-1) 2000+12001=1= 2.

So the answer is: 2.

4. Let α, β be rational numbers and γ be irrational numbers. If α+βγ=0, it proves that α = β = 0.

Proof: Assuming β≠0,

∵α+βγ=0,( 1)

∴γ=- αβ,

And beta are rational numbers,

∴ γ is a rational number, which contradicts that γ is an irrational number.

The assumption of ∴ is not valid.

∴β=0.

Substituting (1), α=0,

∴α=β=0.

5. Prove that 3 is an irrational number.

Proof: Suppose 3 is a rational number.

∫ 1 < 3 < 2, ∴ 3 is not an integer,

Then there are two coprime positive integers p, q, so that 3= pq,

So p = 3q.

Square both sides to get p2=3q2. ..

3q2 is a multiple of 3,

P2 is a multiple of 3,

And ∵p is a positive integer,

P is a multiple of 3.

Let p=3k(k is a positive integer) and substitute it into the above formula to get 3q2=9k2.

∴q2=3k2，

Similarly, q is a multiple of 3,

This contradicts the assumption that P and Q are coprime.

Suppose 3 is a rational number.

So 3 is an irrational number.

6. If two real numbers A and B make a2+b and a+b2 rational numbers, then this pair of numbers (a, b) is harmonic.

① Try to find a pair of irrational numbers to make (a, b) harmonious;

② Prove that if (a, b) is harmonic and a+b is a rational number not equal to 1, then both A and B are rational numbers;

③ Prove that if (a, b) is harmonic and ab is rational, then A and B are rational;

Solution: ① Suppose a= 2+ 12 and b= 12- 2, then A2+B = (2+12) 2+12 =12.

A+B2 = 2+12+(12-2) 2 =114 is a rational number.

Therefore, (a, b)=( 2+ 12, 12- 2) is harmonious;

② it is known that t = (a2+b)-(a+B2) = (a-b) (a+b-1) is a rational number, and a+b=s is a rational number.

So a-b= ta+b- 1, and a= 12(s+ ts- 1) are rational numbers.

Of course, b=s-a is also a rational number;

③ If a+b2=0, then b=- ab is a rational number, and a=(a+b2)-b2 is also a rational number.

If a+b2≠0, it is known that x= a2+ba+b2= (ab)2+ 1bab? 1b+ 1 is a rational number, and y= ab is also a rational number.

So 1b= y2-xxy- 1, so b= xy- 1y2-x is a rational number.

So a=(a+b2)-b2 is also a rational number.