answer:Since e= dfrac {c}{a}= dfrac {2 sqrt {3}}{3}, we have c= dfrac {2 sqrt {3}}{3}a. Therefore, the slope of the asymptotes in the first and third quadrants is dfrac {b}{a}= dfrac { sqrt {c^{2}-a^{2}}}{a}= dfrac { sqrt {3}}{3}.Thus, the inclination angle of this asymptote is 30^{circ}.Therefore, the angle between the two asymptotes of the hyperbola is 2 times 30^{circ}=60^{circ}, which is dfrac {pi}{3}.Hence, the correct choice is boxed{C}.From the eccentricity, we can determine c= dfrac {2 sqrt {3}}{3}a, which allows us to find dfrac {b}{a}= dfrac { sqrt {3}}{3}. Thus, the inclination angle of one asymptote is 30^{circ}, from which we can determine the angle between the two asymptotes.This question examines the standard equation of a hyperbola and the application of simple properties of a hyperbola. The key and difficulty in solving the problem lie in determining that the inclination angle of the asymptotes in the first and third quadrants is 30^{circ}.

answer:1. Based on the identity a^{3}+b^{3}=(a+b)left(a^{2}-a b+b^{2}right), we can factorize the left side, and thus we are left with the equation(2 x-4)left((x-7)^{2}-(x-7)(x+3)+(x+3)^{2}right)=278(x-2) .One solution is obviously x_{1}=2. We will seek the other solutions after canceling both sides by 2(x-2): on the left side, we then have left(x^{2}-14 x+right. 49) -left(x^{2}-4 x-21right)+left(x^{2}+6 x+9right)=x^{2}-4 x+79, and on the right side, we have 139. Therefore, we need to solve the quadratic equation x^{2}-4 x-60=0, and its solutions are x_{2 / 3}=frac{4 pm sqrt{16+240}}{2}=frac{4 pm 16}{2}=2 pm 8, i.e., x_{2}=-6 and x_{3}=10.

answer:(I) From the given function, we have f'(x)=3x^{2}-3. Solving for f'(x)=0 yields x=pm1.When xin(-1,1), f'(x) 0.Therefore, the function f(x) is decreasing on (-1,1) and increasing on (-infty,-1), (1,+infty).When x=-1, the function attains its maximum value, f(-1)=3. When x=1, the function attains its minimum value, f(1)=-1.(II) From f'(x)=3x^{2}-3, we have f'(0)=-3.Since f(0)=1, the equation of the tangent line to the curve at the point (0,f(0)) is given by y-1=-3x, which simplifies to boxed{3x+y-1=0}.

answer:From a^{2}+b mid a^{2} b+a it followsa^{2}+b midleft(a^{2} b+aright)-bleft(a^{2}+bright)=a-b^{2}From b^{2}-a mid a b^{2}+b it followsb^{2}-a midleft(a b^{2}+bright)-aleft(b^{2}-aright)=b+a^{2} .We see that a^{2}+bleft|a-b^{2}right| a^{2}+b. This means that a^{2}+b is equal to a-b^{2} up to a sign. Therefore, we have two cases: a^{2}+b=b^{2}-a and a^{2}+b=a-b^{2}. In the second case, a^{2}+b^{2}=a-b. But a^{2} geq a and b^{2} geq b>-b, so this is impossible. Therefore, we must have the first case: a^{2}+b=b^{2}-a. This gives a^{2}-b^{2}=-a-b, so (a+b)(a-b)=-(a+b). Since a+b is positive, we can divide by it and get a-b=-1, so b=a+1. All pairs that can satisfy the conditions are thus of the form (a, a+1) for a positive integer a.We check these pairs. We have a^{2}+b=a^{2}+a+1 and a^{2} b+a=a^{2}(a+1)+a= a^{3}+a^{2}+a=aleft(a^{2}+a+1right), so the first divisibility relation is satisfied. Furthermore, b^{2}-a=(a+1)^{2}-a=a^{2}+a+1 and a b^{2}+b=a(a+1)^{2}+(a+1)=a^{3}+2 a^{2}+2 a+1= aleft(a^{2}+a+1right)+a^{2}+a+1=(a+1)left(a^{2}+a+1right), so the second divisibility relation is also satisfied. The pairs (a, a+1) thus satisfy the conditions and are therefore the exact solutions.

answer:Answer: (2 ; 2 sqrt{3}),(-4 ; 0),(2 ;-2 sqrt{3}).Solution. Let's denote the points where the ant and the beetle are located as M(alpha) and N(beta) respectively, where alpha and beta are the angles that the radius vectors of points M and N form with the positive direction of the x-axis. Note that the angle between overrightarrow{A M_{0}} and overrightarrow{A N_{0}} is frac{pi}{2}, and at this point alpha_{0}=frac{2 pi}{3}, beta_{0}=frac{pi}{6} - are the angles corresponding to the initial positions of the insects.The distance between the ant and the beetle will be the smallest when the angle between the vectors overrightarrow{A M} and overrightarrow{A N} is zero. Since left|A M_{0}right|=2 and left|A N_{0}right|=4 are the radii of the circles, and left|A N_{0}right|=2left|A M_{0}right|, the angular velocity of the ant is twice the angular velocity of the beetle.Let at the moment of the vectors overrightarrow{A M} and overrightarrow{A N} aligning, the beetle has moved by an angle omega. Then alpha_{0}-2 omega=beta_{0}+omega+2 pi n, where n in mathbb{Z}. Therefore, omega=frac{alpha_{0}-beta_{0}}{3}-frac{2 pi n}{3}=frac{pi}{6}-frac{2 pi n}{3}, where n=0,-1,-2 ldotsThere will be three different points (for n=0,-1,-2). For n=0 we get beta_{1}=beta_{0}+frac{pi}{6}=frac{pi}{3}. The coordinates of the beetle's position are found using the formulas x_{N}=4 cos beta_{1}=2, y_{N}=4 sin beta_{1}=2 sqrt{3}.The other points are obtained by rotating point N_{1} around the origin by angles frac{2 pi}{3}, frac{4 pi}{3} and have, respectively, coordinates (-4 ; 0) and (2 ;-2 sqrt{3}).

answer:To solve this problem, let's analyze the properties of both squares and rectangles and compare them to identify which property is unique to a square.1. Four right angles: Both squares and rectangles have this property. A square is a special type of rectangle, but this doesn't distinguish a square from a rectangle.2. Opposite sides are parallel and equal: This is another common property shared by both squares and rectangles. It indicates that both shapes have their opposite sides equal in length and parallel to each other.3. Diagonals bisect each other: This property is also shared by both squares and rectangles. It means that the diagonals of both shapes cut each other exactly in half.4. Diagonals are perpendicular to each other: This is where squares and rectangles differ. While the diagonals of a rectangle bisect each other and are equal in length, they are not necessarily perpendicular. In contrast, the diagonals of a square not only bisect each other but are also perpendicular to each other. This is a unique property of squares that rectangles do not possess.Therefore, by analyzing the properties listed above, we can conclude that the property unique to a square, which a rectangle does not have, is that its diagonals are perpendicular to each other.Hence, the correct answer is boxed{D}.