answer:begin{array}{l}sin alpha cdot cos beta + cos alpha cdot sin beta = 3(sin alpha cdot cos beta - cos alpha cdot sin beta), sin alpha cdot cos beta = 2 cos alpha cdot sin beta. text{Therefore, } frac{tan alpha}{tan beta} = frac{sin alpha cdot cos beta}{cos alpha cdot sin beta} = 2.end{array}

answer:4. sqrt{3}+1.As shown in Figure 2, with BC as the x-axis and the perpendicular bisector of BC as the y-axis, establish a coordinate system. Suppose the equation of the hyperbola is frac{x^{2}}{a^{2}}-frac{y^{2}}{b^{2}}=1, then C(c, 0), A(0, sqrt{3} c), so Eleft(frac{c}{2}, frac{sqrt{3}}{2} cright). Substituting into the hyperbola equation givesfrac{c^{2}}{4 a^{2}}-frac{3 c^{2}}{4 b^{2}} datawhich is frac{c^{2}}{4 a^{2}}-frac{3 c^{2}}{4left(c^{2}-a^{2}right)}=1 text {, }which is frac{e^{2}}{4}-frac{3 e^{2}}{4left(e^{2}-1right)}=1.Solving for e^{2} gives e^{2}=4 pm 2 sqrt{3}.begin{array}{l}because e>1, therefore e=sqrt{4+2 sqrt{3}}=end{array}sqrt{3}+1

answer:(I) Since p wedge q is true, both p and q are true.Given m=1, p is true when left|2x-1right|geqslant 1, which implies xleqslant 0 or xgeqslant 1.For q to be true, we need dfrac{1-3x}{x+2} > 0, which implies (1-3x)(x+2) > 0, hence -2 < x < dfrac{1}{3}.Combining these two conditions, we get -2 < xleqslant 0.Thus, the range of the real number x is (-2,0].(II) For p, the real number x satisfies left|2x-mright|geqslant 1, hence neg p implies left|2x-mright| < 1, which gives dfrac{m-1}{2} < x < dfrac{m+1}{2}. Let A=(dfrac{m-1}{2}, dfrac{m+1}{2}).For q, we have -2 < x < dfrac{1}{3}. Let B=(-2, dfrac{1}{3}).Since neg p is a sufficient but not necessary condition for q, A is a proper subset of B.This gives us begin{cases} dfrac{m-1}{2}geqslant -2 dfrac{m+1}{2}leqslant dfrac{1}{3}end{cases} (Note: Equality cannot hold simultaneously). Solving these inequalities, we get -3leqslant mleqslant - dfrac{1}{3}.Thus, the range of the real number m is boxed{[-3,- dfrac{1}{3}]}.

answer:To solve this problem, we follow a step-by-step approach based on the given solution.Step 1: Analyzing the Linear FunctionGiven the linear function y=(7-a)x+a, for it not to pass through the fourth quadrant, two conditions must be met:1. The slope of the line, which is (7-a), must be positive. This ensures the line is increasing and not descending into the fourth quadrant.2. The y-intercept, which is a, must be non-negative. This ensures the line does not start below the x-axis.Thus, we have:[left{begin{array}{l}7-a > 0 a geq 0end{array}right.]Solving these inequalities, we find:[0 leq a < 7]Step 2: Solving the Fractional EquationGiven the fractional equation frac{6x}{x-1}=3+frac{ax}{x-1}, we simplify it to find the value of x in terms of a:[frac{6x}{x-1} - frac{ax}{x-1} = 3][frac{6x - ax}{x-1} = 3][frac{(6-a)x}{x-1} = 3]Setting the numerators equal since the denominators are the same, we get:[(6-a)x = 3(x-1)]Solving for x, we find:[x = frac{3}{a-3}]Step 3: Ensuring x is an IntegerFor x to be an integer, and given x neq 1 to avoid division by zero in the original equation, we examine the denominator a-3. The numerator 3 is fixed, so for x to be an integer, a-3 must be a divisor of 3. Considering the range of a from Step 1 (0 leq a < 7), the possible values of a that make x an integer are a=0, 2, and 4.Conclusion:The number of integers a that satisfy all conditions is 3. Therefore, the correct answer is encapsulated as follows:[boxed{C}]

answer:For the first part, here is one maximal arrangement, where the location of the bishops are indicated by the letter B.| B | | | | | | | || :---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- || B | | | | | | | B || B | | | | | | | B || B | | | | | | | B || B | | | | | | | B || B | | | | | | | B || B | | | | | | | B || B | | | | | | | |To see that there cannot be 15 bishops, observe that we have highlighted 15 right-down diagonals in the square above. Each diagonal can accommodate at most one bishop. Furthermore, the lower-left corner and the upper-right corner constitute diagonals of size 1 which cannot be both occupied. This gives the bound of 14 bishops.

answer:begin{array}{l}text { Sol } quad text { Let the original expression }=x, quad text { then } x^{2}=8+ 2 sqrt{10+2 sqrt{5}}+2 sqrt{64-4(10+2 sqrt{5})} +8-2 sqrt{10-2 sqrt{5}}, x^{2}=16+4(sqrt{5}-1) quad=(sqrt{10}+sqrt{2})^{2} . because x>0, therefore x=sqrt{10}+sqrt{2} .end{array}Note. This method is applicable to sqrt[n]{A+sqrt[n]{B}} pm sqrt[n]{A-sqrt[n]{B}}.