This question is an extension of the classical question type. A point P, PA=3, PB=4 and PC=5 of the equilateral triangle ABC rotates, and rotation △ABP makes AB and BC coincide. After the rotation, the point P is marked as P', and it is easy to prove that △P'Pb is an equilateral triangle, so P'P = Pb = 4, P'C = 3, PC.
3. Algebraic deformation? (x-a+a)? \(x-a)=(x-a)+a? \(x-a)+2a=[ followed by (x-a)]? -[a\ followed by (x-a)]? +4a? So the minimum value is 4a.
4. Because the length of AB is a fixed value? Therefore, the smallest perimeter is the smallest of the other three sides. Let a be the symmetry point E(-8. -3), and b is the symmetrical point f (4,5) about the y axis. For any point C and D, BC+CD+AD=CF+CD+ED, and the distance between E and F is the shortest, that is, the straight line segment of EF is the shortest. At this time, the EF linear equation is obtained. The coordinates of c and d are (0,7 \ 3) and (-7 \ 2,0) respectively, that is, m = 7 \ 3 and n =-7 \ 2, so the value of m: n is -2\3.
5. from the meaning of the question: △=b? -20c & gt; =0, sum range of two roots: -2
1 Annual Counseling Work Summary
This semester, I am honored to be Wang's teacher. He is very experienced in the management of the head tea