answer:Solve: From x=frac{sqrt{3}-1}{sqrt{3}+1}=2-sqrt{3}, we get x-2=sqrt{3}. Squaring it, we obtainx^{2}-4 x+1=0 text {. }Use equation (1) to reduce the power of the target algebraic expression. Performing polynomial division, we getbegin{array}{l}x^{4}-5 x^{3}+6 x^{2}-5 x+4 =left(x^{2}-4 x+1right)left(x^{2}-x+1right)+3=3 .end{array}

answer:23. Since a square can end with the digits 0,1,4,5,6 or 9, and each subsequent digit of a six-digit number is greater than the previous one, a six-digit number that is the square of a three-digit number can end with either the digit 6 or 9. If it ends with 6, it would be written as 123456. But this number is not a perfect square. Therefore, it must end with the digit 9, which means the sought three-digit number ends with the digit 3 or 7. The leftmost digit of the six-digit number cannot be 4, as the number 456789 is not a perfect square. Therefore, it is less than 4. Let's consider 1 * * * * 9, then the three-digit number is either 4 * 7 or 3 * 7. But 407, 417,427,437,447 do not satisfy the condition, and 457^{2}=208849. (We get a number with the leftmost digit not 1, but 2.) We will look for numbers of the form 3 * 7. Checking shows that such a number is 367left(367^{2}=134689right).

answer:Sol: because 0<log _{5} 3<1<log _{2} 3,therefore y_{1}=left(log _{2} 3right)^{2} is an increasing function.Also, becauseleft(log _{5} 3right)^{x} is a decreasing function,therefore y_{2}=-left(log _{5} 3right)^{prime} is an increasing function.Therefore, y=y_{1}+y_{2}=left(log _{2} 3right)^{prime}-left(log _{5} 3right)^{prime} isan increasing function.Given left(log _{2} 3right)^{x}-left(log _{5} 3right)^{x} geqslantleft(log _{2} 3right)^{-v}-left(log _{5} 3right)^{-y}, we get x geqslant-y, i.e., x+y geqslant 0. Therefore, the correct choice is(B).

answer:1. To have as few angles as possible, their measures need to be as large as possible. We can take at most two angles of measure 125^{circ} (and we will take exactly two!), at most four angles of measure 25^{circ} (and we will take exactly four), and, finally, two more angles of measure 5^{circ}. The minimum number of angles under the conditions of the problem is, therefore, eight.

answer:According to the problem, a=2, a+c=2+sqrt{3}, then c=sqrt{3}, b=1. Let angle E P F=theta,|P E|=x,|P F|=y, in triangle P E F, by the cosine rule we get 4 c^{2}=x^{2}+y^{2}-2 x y cos theta=(x+y)^{2}-2 x y(1+cos theta).Thus 12=16-4(1+cos theta) Rightarrow cos theta=0 Rightarrow theta=90^{circ}, so S_{triangle P E F}=b^{2} tan frac{theta}{2}=1.

answer:Answer: 2.5 km/h.Solution. Consider the reference frame associated with the river. In this frame, the boats move with equal speeds: initially, they move away from each other in opposite directions for 1 hour, then move in the same direction for another 1 hour (during which the distance between them does not change), and then approach each other until they meet. Since the distance between the boats did not change during the second time interval, they will approach each other for as long as they moved away, which is 1 hour. Thus, from the moment of departure until the meeting, 3 hours have passed. During this time, the log (moving with the current) traveled 7.5 km. Therefore, its speed is 2.5 km/h.