Compare the size of a 5 b 5 with a 3b 2 a 2b 3 10

Subtract the two formulas.

a^5+b^5)-(a^3b^2+a^2b^3)a^3(a^2-b^2)+b^3(b^2-a^2)(a^3-b^3)(a^2-b^2)

a-b)(a 2+b 2+ab)(a-b)(a+b)(a-b) 2(a 2+b 2+ab)(a+b) (a+b) (a-b) (a-b) 2 is positive because a is not equal to b.

And (a 2 + b 2 + ab) is always positive, so the result of subtracting two formulas depends on the size of a+b.

If a+b is positive, then the preceding is greater than the lasting.

If it is negative, the front is smaller than the back.

If it is 0, it is equal.

a 5+b 5-a 3b 2-a 2b 3(a-b) 2(a 2+ab+b 2)(a+b) where (a-b) 2 (a 2+ab+b 2) are all greater than 0, so when a+b is greater than 0, the original formula is greater than 0

When a+b is less than 0, the original formula is less than 0

Just find a number to substitute in...

Although this method is a bit stupid. But it's intuitive.

First, we can calculate the difference between 2a+3b and a-2b+3, i.e., (2a+3b)-(a-2b+3)=a+5b-3. If this difference is greater than 0, then Chunpei 2a+3b is greater than a-2b+3;If this difference is less than 0, then 2a+3b is less than a-2b+3.

Because the hole envy is a>b>0, a+5b-3=a+2b+3b-3>a+2b>0, so 2a+3b is greater than a-2b+3.

Therefore, 2a+3b is greater than a-2b+3.

The comparison size is the two numbers of loss and the remainder of the phase minus the pin, the difference between the years is a positive number, the number is large, and the difference is negative and the back is large.

2a+4-(2b-5)

2a-2b+9

2(a-b)+9

So 2a+4 is bigger.

Knowing that 5a-3b-2=5b-3a, use the properties of the equation to compare the magnitude of a and b?

5a 3b 2=5b 3a Move a and b to different sides to get 8a 2=8b, and it can be concluded that a is greater than b