Math proof questions, ask for solutions

Consider assuming that the sum of all x i is positive. Denote the part inside the absolute sign as f(k), then f(0)<-1, f(n)>1Note that f(k)-f(k-1)=2x k, so |f(k)-f(k-1)|<=2, so when k gradually increases from 0 to n, the change of f(k) in each step does not exceed 2, and it cannot always be outside the range of length 2 [-1,1].

bpc＝150°

In fact, the question gave the auxiliary line to which Tong came out, because there was no need for Li Zhentan to give the E point the number of trips.

(1) F is the midpoint of AC and E is the midpoint of Cd, then EF AD, EF=AD 2, AC=AD, EF=AC2;

2) In the right-angle ABC, AC is the hypotenuse, then AF=FC=BF=AC 2, then BF=EF, BFE is isosceles, G is the midpoint of the bottom edge BE, then FG BE.

AF bisects DFE and CD AB, be AC

adf＝∠aef，∠daf＝∠eaf

and af for the common side.

ADF is equal to AEF

AD AE is in ADC and AEB.

adf＝∠aef｝

ad＝ae｝

a a (Public Corner)

ADC Quanxiang Wide is purely equal to Qiaopai AEB

ab ac (congruent triangles correspond to equal sides).

(1) Connect AC to prove that the triangle ABE and triangle ACF are congruent (2) Since AB is parallel to CD, the angle BAF + angle CFA is 180 degrees.

i.e., angle BAE + 60 + angle CFE + 60 180 so angle CFE 45 degrees.

1. The conditions do not seem to be enough, you can draw an unequal triangle,? , prove that the three angles of AEF are 60 degrees, or AE=AF isosceles.

2.According to the conclusion of question 1, the angle AEB = 180-60-15 = 105, the angle c = 180-60 = 120, the angle fec = 180-105-60 = 15, and the angle CFE = 180-120-15 = 45

Question 1: Certificate: Connecting ac b=60°, abc is an equilateral triangle, c=120°; ∵eaf=60°，∴eaf＋∠c=180°；∴aec+∠afc=180°；afd afc=180°, aec= afd, acb= d, ac=ad, ace adf, ae=af, eaf=60°, aef is an equilateral triangle.

Question 2: It is known that AEB=105°, AEC+ AFC=180°, and AEB+ AEC=180°, so AFC= AEB=105°, and AFE=60°, CFE=105°-60°=45°.