It is known that positive numbers a, b satisfy a3b ab3 2a2b 2ab2 7ab 8, a2 b2 15

Firstly, a3b+ab3-2a2b+2ab2=7ab-8 is transformed into ab(a-b-1)2+2(ab-2)2=0 by extracting the common factor, using the perfect flat method, and adding terms to obtain a-b=1 and ab=2 according to the properties of a and b being positive and non-negative numbers, and then solving the values of a and b, and substituting a2-b2 to obtain the result

a3b+ab3-2a2b+2ab2=7ab-8，ab（a2+b2）-2ab（a-b）=7ab-8，ab（a2-2ab+b2）-2ab（a-b）+2a2b2-7ab+8=0，ab（a-b）2-2ab（a-b）+2a2b2-7ab+8=0，ab[（a-b）2-2（a-b）+1]+2（a2b2-4ab+4）=0, ab(a-b-1)2+2(ab-2)2=0, a, b are positive, ab 0, a-b-1=0, ab-2=0, that is, a-b=1, ab=2, solve the equation ab1ab2, solve a=2, b=1, a=-1, b=-2 (not in line, rounded), a2-b2=4-1=3

Therefore, choose B

Problem solving idea: firstly, a 3b+ab 3-2a 2b+2ab 2=7ab-8 is converted into ab(a-b-1) 2+2(ab-2) 2=0 by extracting the common factor, using the perfect flat method, and adding terms into ab(a-b-1) 2+2(ab-2) 2=0, and then according to the properties of a and b are positive numbers and non-negative numbers, a-b=1 and ab=2 are obtained, and then the values of a and b are solved, and the results are obtained by substituting a 2-b 2

a3b+ab3-2a2b+2ab2=7ab-8, ab(a2+b2)-2ab(a-b)=7ab-8, ab(a2-2ab+b2)-2ab(a-b)+2a2b2-7ab+8=0, pat trapped nucleus.

ab(a-b)2-2ab(a-b)+2a2b2-7ab+8=0,ab[(a-b)2-2(a-b)+1]+2(a2b2-4ab+4)=0, ab(a-b-1)2+2(ab-2)raid 2=0, a, b are positive, ab 0, a-b-1=0, ab-2=0, i.e. a-b=1, ab=2, solve the equation.

a b 1ab 2, solution a=2, b=1, a=-1, b=-2 (not in line with the topic, discarded), a2-b2=4-1=3

Therefore, b 2 is selected, and the positive number a is known, and b satisfies a 3b + ab 3-2a 2b + 2ab 2 = 7ab-8, then a 2-b 2 = (

a. 1 b. 3

c. 5 d.Not sure.

7.Knowing that the positive number a, b satisfies 3a2+14ab+8b2=25, then the final answer of 2a+3 is 5

3a2+14ab+8b2=25 can be reduced to (a+4b)(3a+2b) 25

then a 1 b 1

Then 2a+3 5

You're wrong about that.

a=3c, b=2c, with a bridge into the world AC+BC=AB: 3C2+2C2=6C2, and a, b, and c are all positive numbers.

Approximate distribution: 3+2=6, i.e. 5=6;Obviously, the elder brother is not established.

a+2ab+b=16 a+b=4,ab=4 a=b=2 This is mainly the application of the perfect sum of squares formula.

The above formula can be transformed into a code bridge.

a-1 2b) 2+3 4(b-2) 2+(c-1) 2 0

Therefore a=1 2b b=2 c=1

i.e. a=1, b=2, c=1

This is the domain of linear programming.

With a,b as the coordinate axis, then 20,d represents the distance from the origin (0,0) to the point (a,b) in the "belt" region.

Since the distance from the origin to the straight line a+2b-2=0 is.

d1=|0+0-2|(1 +2 )=2 5 The distance from the origin of the line a+2b-4=0 is.

d2=4/√5

Thus, 2 54 5 i.e. a +b can be taken in the range of (4 5, 16 5).

With a,b as the coordinate axis, then 20,d represents the distance from the origin (0,0) to the point (a,b) in the "belt" region.

Since the distance from the origin to the straight line a+2b-2=0 is.

d1=|0 0-2|(1 +2 )=2 5 The distance from the origin of the line a+2b-4=0 is.

d2=4/√5

Thus, 2 54 5 i.e. a +b can be taken in the range of (4 5, 16 5).

If a,b are positive integers, the answer is 2