If you are anxious to solve a junior high school math problem, be sure to try to be as detailed and

The picture is not easy to draw, so it is not drawn. Thank you for your understanding.

Solution: (1) The x-axis and y-axis are regarded as mirrors, and p and q are regarded as point light sources. Follow the light and take shortcuts to draw (do it yourself).

Method: Make the image point p in the y-axis mirror'The image point q with q in the mirror surface of the x-axis'。Connect p'q'The x and y axes are intersected at two points, m and n.

At this time, the quadrilateral PQMN has the smallest circumference. In fact, the principle is very simple, and it is easy to know p from the diagram'n=pn、q'm=qm, so the perimeter of the quadrilateral pqmn=pq+pn+mn+qm=pq+p'n+mn+q'm=pq+p'q'。The length of pq has been determined, and at this time p'q'is a line segment, and the line segment within two points is the shortest, so the quadrilateral PQMN circumference is the shortest.

2) The coordinates of m and n can be found analytically: m(1,0); n(0,1)。

a^3+a^2=6a+6

a^2(a+1)=6(a+1)

a^2-6)(a+1)=0

a + root number 6) (a - root number 6) (a + 1) = 0a1 = - root number 6, a2 = root number 6, a3 = -1

Root number 6 So the closest approximation is c

to (a 2-6)(a+1)=0

The root is -1, 6

Because 6 is approximately equal to, to be exact, that is.

Therefore, the answer is c

Hope it helps.

Select c to simplify the original equation: a 2 (a + 1) = 6 (a + 1).

Simplified: a 2 = 6

The easiest way to do this is to bring in the calculation and see which one is closest to 6 squared.

The original solution gives a 2 = 6, that is, a = root number 6, and the approximate value is closest to c, choose c

With d as the origin and the straight line where AD is located as the x-axis, a Cartesian coordinate system is constructed.

a(-3,0), c(2,-1), easy to get e(1,2), then.

ae=√｛（3-1）²+0-2）²｝=2√5.

Analysis: According to the question, if EF AD is F, DG BC is G, and it is proved that CDG is equal to EDF, then the value of AE can be obtained

Answer: Solution: As shown in the figure, as EF AD is f, dg bc is in g, according to the nature of rotation, we can know that de=dc, de dc, cdg= edf, cdg edf, df=dg=1, ef=gc=2, ae= root number (16+4) = 2 root number 5

It is known that in the trapezoidal ABCD, AD BC, B=90, AD=3, BC=5, AB=1, the line CD is rotated 90 to the DE position around the point D, and AE is connected, then the length of AE is (2 5).

Pass E for EH AD, cross AD extension line at H, pass D for DF BC at F

EH=fc=BC-AD=5-3=2DH=DF=AB=1 from known demonstrable EDH CDF

In the right triangle EAH, EH=2, Ah=AD+DH=3+1=4, so AE= (2 2+4 2)=2 5

Solution 2: First, find that DC length is equal to root number 5 and AC is equal to root number 26, and then use the triangle cosine theorem to find COS ADC

, so that the value of the sin adc can be obtained, and then the ae value can be found using the formula of the triangle cosine theorem (you should know). In the process of bounding, it is necessary to convert cos ade to sin adc [cos( ade)=cos(2 -1 2 - adc)=sin adc] to solve that ae is equal to the root number 20

A to B is the water, and the time t1=s (a+b).

From b to a is reverse water, time t2=s (a-b).

Average velocity = 2s (t1+t2) = 2s [s (a+b)+s (a-b)] = (a 2-b 2) a

First calculate the time it takes to go against the water, then calculate the time it takes to go against the water, and finally use the total distance to remove the sum of the two times. The speed of the river is equal to the speed of the ship plus the speed of the water flow a+b, and the speed of the reverse water is equal to the speed of the ship minus the speed of the water a-b, and the total distance should be 2skm for one round trip.