answer:By the Cosine Rule, we have 12=b^{2}+c^{2}-bc.Also, the area of the triangle is given by S= dfrac {1}{2}bcsin A= dfrac { sqrt {3}}{4}bc.Now, using the inequality b^{2}+c^{2}geqslant 2bc, we can rearrange to get bc+12geqslant 2bc, which simplifies to bcleqslant 12 (with equality holding if and only if b=c).Therefore, the maximum area S is achieved when b=c, and is given by S= dfrac {1}{2}bcsin A= dfrac { sqrt {3}}{4}bc leqslant 3 sqrt {3}.Hence, the maximum value of the area S of triangle ABC is boxed{3 sqrt {3}}.

answer:S o l u t i o n. Several methods can be proposed to solve the given equation.1st method. Since the equation is quadratic (if we do not consider x under the cosine sign), we can express x in terms of trigonometric functions of the angle x y:x=-2 cos x y pm sqrt{4 cos ^{2} x y-4}=-2 cos x y pm 2 sqrt{-sin ^{2} x y}Since by condition x and y are real numbers, then -sin ^{2} x y geqslant 0. This inequality is only satisfied when sin x y=0, i.e., when x y=k pi, k in mathbf{Z}.Now we have:x=-2 cos x y=-2 cos k pi .If k is even, then cos k pi=1 and x=-2, so y=n pi, n in mathbf{Z}; if k is odd, then cos k pi=-1 and x=2, so y=frac{pi}{2}(2 m+ +1), where m in mathbf{Z}.2nd method. Transform the equation so that the left side becomes a sum of non-negative numbers, for example by adding and correspondingly subtracting 4 cos ^{2} x y. We get:(x+2 cos x y)^{2}+4left(1-cos ^{2} x yright)=0or(x+2 cos x y)^{2}+4 sin ^{2} x y=0Since the sum of squares of real numbers equals zero only when each term is zero, we arrive at the system:left{begin{array}{c}x+2 cos x y=0 sin x y=0end{array}right.Solving this system, we find the same values of x and y.

answer:【Solution】Solution: 80 times 4-4 times 60=80 (people), that is, x hours after the payment starts, there is no customer in the queue. According to the problem, we can get the equation:begin{aligned}80 times 2 times x & =80+60 x, 100 x & =80, x & =0.8,end{aligned}Answer: There will be no queue 0.8 hours after the payment starts. Therefore, the answer is: 0.8.

answer:19. From the conditions, when m mid n and m < n, we have a_{n} geqslant 2 a_{m}. Therefore, a_{1} geqslant 1, a_{2} geqslant 2, a_{4} geqslant 2 a_{2} geqslant 2^{2}, similarly, a_{8} geqslant 2^{3}, a_{16} geqslant 2^{4}, a_{30} geqslant 2^{5}, a_{400} geqslant 2^{6}, a_{2000} geqslant 2^{7}, which means a_{2000} geqslant 128.On the other hand, for any positive integer n, let the prime factorization of n ben=p_{1}^{q_{1}} p_{2}^{q_{2}} cdots p_{k^{prime}}^{q_{k}}where p_{1}<p_{2}<cdots<p_{k} are prime numbers, and alpha_{1}, alpha_{2}, cdots, alpha_{k} are positive integers. Definea_{n}=2^{a_{1}+a_{2}+cdots+a_{k}}Then the sequence left{a_{n}right} satisfies the requirements of the problem, anda_{2000}=2^{4+3} leqslant 2^{7}Therefore, the minimum value of a_{2000} is 128.

answer:Prove that by the intersecting chords theorem, we have ( E C cdot C F = A C cdot C B ).Since ( P E = P F ), applying the corollary 1 of Stewart's theorem to the isosceles (triangle P E F) and point (C) on the base (E F), we have ( P C^{2} = P E^{2} - E C cdot C F ), which means[begin{aligned}P E^{2} & = P C^{2} + E C cdot C F = P C^{2} + A C cdot C B & = P C^{2} + (P C - P A) cdot (P B - P C) & = P C^{2} - P C^{2} - P A cdot P B + P C cdot P B + P C cdot P A & = P A cdot P C + P B cdot P C - P A cdot P B .end{aligned}]Since ( P E^{2} = P A cdot P B ), it follows that ( 2 P A cdot P B = P A cdot P C + P B cdot P C ).Thus, (frac{2}{P C} = frac{1}{P A} + frac{1}{P B}).

answer:Solution. It is easy to notice that x=frac{1}{3} is a solution. For x<-frac{1}{3}, the left side of the equation is negative, while the right side is positive, so there can be no roots. For x geq-frac{1}{3}, the left side monotonically increases, therefore, the equation can have only one root, so there are no other roots.