22. The score of the suggested topic can actually be expressed by (A 3+B 3)/(A 3+C 3), where A = B+C.

The proof process shows that the cubic sum formula A 3+B 3 = (A+B) (A 2-AB+B 2), such items as A 2-AB+B 2 can be replaced by method of substitution, and A 2-AB+B 2 = A 2-AC can be obtained.

8. Tip: In the substitution method, y 2 = (-3/2) x 2+3x is obtained from 3x 2+2y 2-6x = 0. Replace it.

X 2+Y 2 can give a univariate quadratic function about X, and the value range of X should be determined by Y 2 = (-3/2) x 2+3x, that is, (-3/2) x 2+3x ≥ 0.

13, prompt, ① question, use the relationship between △ and 0 to judge the intersection. Delta = 4 (b 2-AC) is finally obtained.

∫a+b+c = 0, a>b>c This shows that the largest A must be greater than 0, and the smallest C must be less than 0, so b 2-AC > 0, so △ > 0 。 So there are two different intersections.

(2) Question, the long mathematical expression of this projection is the value of the root sign (XA-XB) 2 (why do you want to be clear).

The radical sign is expanded into = (xa+XB) 2-4xaxb = (4b 2-4ac)/a 2.

When b=0, a+c=0. The value is 4. B<0, a=-b-c substitution, 4-4bc/(b+c) 2 = 4-4bc/[(2bc) 2] ≥ 3, but < 4; B>0, substituting c=-a-b, and getting 4+[4b (a+b)]/a 2 > 4, then the minimum value of the original formula is root number 3;

Last question: hint, because the inequality x 2+ax+b

Hu ~ ~ ~ finally finished, the detailed process is too long, so it's all some tips that I hope will help you.