The third year of junior high school math problems, please take a look at your face!!

Let's assume that the two roots are x1 and x2 respectively

Rule. x1 2-3mx1+2*(m2+m-2)=0x2 2-3mx2+2*(m2+m-2)=0

x1 2-x2 2-3m(x1-x2)=0=>(x1-x2)(x1+x2)-3m(x1-x2)=0=>(x1-x2)(x1+x2-3m)=0, then x1=x2 or x1+x2=3m

1) If x1 = x2, then the quadrilateral with parallel and equal opposite sides is of course a parallelogram.

2) If x1+x2=3m

First, find mx 2-3mx+2*(m 2+m-2)=0=>(x-3 2m) 2-9 4m 2+2m 2+2m-4=0=>(x-3 2m) 2-1 4m 2+2m-4=0=>(x-3 2m) 2=1 4m 2-2m+4=>(x-3 2m) 2=(1 2m-2) 2=>(x-3 2m)=(1 2m-2).

x=1/2m-2+3/2m=2m-2

Obviously, if there are two real roots, their sum should be 4m-4, and the sum of them is 3m, then m=4

So x1+x2=12

In this case, the shape of the quadrilateral cannot be determined.

The answer is the same as upstairs, but here you can also use a direct triangle instead of a similar triangle to solve it:

Connect OF, OA, because AF is tangent to the semicircle, resulting that AF is perpendicular to OF;

and ob=of, so it is concluded that the triangle abo is all equal to the triangle afo, so ab=af=4

In the same way, it can be concluded that ef=ec

Now let de=x, then, ec=4-x=ef

Triangle ade:

ad=4de=x

ae=af+ef=4+4-x=8-x

According to the Pythagorean theorem, the equation is derived:

4x4+xx=(8-x)(8-x)

Solve the equation and get x=3

Finally, the triangle ade=4 x 3 x 6

Quadrilateral ABOF Quadrilateral ocef

s quadrilateral abof=ob ab=8

oc/ab=1/2

S quadrilateral ABOF = 4S quadrilateral ocef

s quadrilateral abce = 8 + 2 = 10

s△ade=16-10=6

According to the inscription, we can know that the angle mon is a right angle, the triangle mon is a right-angled isosceles triangle, do o to mn the vertical line intersection with mn intersection at the point p, op = root number 2, sin angle noe = sin angle nob = sin (angle pon plus angle pob), sin angle pob = 4 points of the root number 14, cos angle pon = 4 points of the root number 2, sin angle pon = 2 points of the root number 2, cos angle pon = 2 points of the root number 2, you can find the sin angle noe =1/4 + root number 7, it should be right, I checked it twice.

45° connection BM, BN, OM, ON, connection MN, BE to F. Because bm, bn is the radius of the circle b, om, on is the radius of the circle o, so bm=bn, om, on, and because bno and bmo have the same side, so bno bmo, so nbo= mbo, so be is the mn perpendicular, so nf=mn 2 = root number 2, so sin noe = the root number 2 of the two, so noe = 45°

Because I don't know the picture, I don't know if it's right, and I can't type too many words on my phone, and the result is twice as many as the root number 5.

Through the O point to do of of perpendicular ad to the point F to connect ot, it is easy to know from the question of=root number 3, because the quadrilateral ofdt is rectangular.

According to the perpendicular theorem, af=1, so the triangle aof, af=1, of=root number 3, angle ofa=90 degrees, so angle aof=30 degrees.

Because the quadrilateral obdt is rectangular, the angle bot=60 degrees, because the triangle obt is an isosceles triangle, so obt is an equilateral triangle, so the angle obt=60 degrees, so the angle abt=60 degrees.

This has something to do with the length of the rectangular pool!

For example, if the inside of a rectangular pool is 4 meters long, i.e. all pools are connected, so just connect p and one side.

And if some parts are connected, there is still uncertainty.

If all of them are not connected, (according to the meaning of the title may be understood in this way) then the shortest is the sum of 6 perpendicular lines: 6 * 2 root number 3 = 12 root number 3 As for the proof, you need to ask, but I guess there is no time today!

On this basis, you can take any cable to connect, and then compare the size (length) of the connected part, so I don't know if you can understand!

This means that there must be pipes from P to the six pools.

Then the vertical line segment is the shortest.

Taking the triangle PAB as an example, the hexagon becomes 4 and the triangle is equilateral, so the height of the AB side is 2 3

So six pipes are 6x2 3 = 12 3

Take a three-shape analysis, such as a triangular PEF, and make a high, pH

Because of PH+PE< pe, so, the answer is 12 * root number 3 + 24 < p>

If the perpendicular segment is the shortest, the shortest route is the sum of the perpendicular segments from p to each side, which can be solved by the known side length and the regular hexagon.

Solution: Connect OA, OF, and OE. Let ce=xcm.

ofe= oce=90°,oe=oe,of=oc,so rt oef rt oec,ef=ec=xcm.

ofa= oba=90°, of=ob, af=ab, so rt oaf rt oab, push af=ab=4cm.

So, ae = (4 + x) cm, ed = (4-x) cm.

From ad 2+ed 2=ae 2: 16+(4-x) 2=(4+x) 2, solution: x=1.

s△ade=1/2*ad*de=1/2*4*3=6(cm^2)