Mathematics training questions, junior high school, junior high school mathematics training exercise

Doudou raised two tanks of goldfish, if one is taken out from the A tank and put into the B tank, then the number of goldfish in the two tanks is equal, if one is taken out from the B tank and put into the A tank, the number of the B tank is 50% of the A tank, how many are there in the two tanks A and B?

Solution: Divide the north quadrilateral into two triangles.

Let the left triangle be placed x only, and the right triangle y only.

According to the ratio of the area of the triangle of the same height to the ratio of the base, it can be obtained:

West + x): y = south: east.

East + y): x = south: west.

That is: (5+x): y=10:8

8+y)：x=10：5

Solution: x=10

y=12, then x+y=10+12=22

So there are 22 sheep in the northern meadow.

1) Xiangli analysis: (a+c)(a+d)=1, (b+c)(b+d)=1 structural equation (x+c)(x+d)=1

Equation, collated: x + (c + d) x + cd-1 = 0 from Vedic theorem, get: Cover Yan Zheng a+b=-(c+d) that is, a+d=-(b+c)Wusong (a+c)(b+c)=-a+c)(a+d)=-1 only know the first -

Take the midpoint O of the CD and connect it to the OAThe intersection is early and round in e

oc=1/2cd=6m，ac=8m。

oa=10m

ae=oa-oe=10-6=4m

The rope should not exceed 4 meters.

1. Counterproof, if two equations do not have unequal real roots. then the discriminant formula is not greater than 0, that is, 1-4b 0, and a -4c 0From these two inequalities, b 1 4, cut c a 4, and bring in the equation a=b+c+1, then, a=b+c+1 1 4+a 4+1, i.e., a 1 4+a 4, collated get:

A -4a+5 0, and then squared to get (a-2) 1, there is a contradiction. So, at least one equation has two unequal real roots.

2.Let the common root of the two equations be c, then we have: (a 1)c (a +2)c+(a +2a)=0, and (b 1)c (b +2)c+(b +2b)=0.

And, (b-c)(b+c+2-bc)=0. <2>

These two equations are discussed below:

1) C is not equal to any of A, B, i.e. C≠A≠B.

1-c)a+(c+2)=0, and: (1-c)b+(c+2)=0, obviously the two equations are not valid when c=1, so c≠1then there is a=b= -(c+2) (1-c).

This contradicts the fact that the coefficients of the first terms of the two equations are not equal, i.e., a≠b. Therefore, this situation is not established.

2) c is equal to one of a and b, you may wish to set c=a

At this point, equation <1> is automatically established. Bringing c=a into the equation<2> yields (b-a)(b+a+2-ab)=0. It is obtained from a≠b, b+a+2-ab=0, and (a-1)(b-1)=3 can be obtained after sorting.

Since a,b are positive integers, we get a=2, b=4, or a=4, b=2. If c=b is assumed, the result is the same.

In summary, the results are a=2, b=4, or a=4, b=2. (Actually, the two equations in the problem have two common roots, but fortunately, the problem only says "there is a common root", not "there is only one common root", otherwise there will be no solution).

3) This is really a mistake in the problem, very simple, the discriminant formula of the root b -4ac 0 can have a real root, a=1, c=3, b=101, obviously can make the equation have a real root...

Wow!!! After writing so much, I found out that your bounty score is 0... Good injury.

In junior high school mathematics, this type of question is tested every year, and you must master it if you want to get a high score.

b-1 solution: |ae|=25 |ae|＝4de＋2de＋4／3de＋de＝25／3de

The solution is: de=3 cd=4 bc=6 ab 12, so point d represents 9, point c represents 5, point b represents 1ae, and the midpoint of 1ae is ( 13 12) 2 0 5, so point b-1 is the closest.

1. Substitute the points a(4,0) and b(1,3) into y=-x2+bx+c.

0=16+4b+c

3=1+b+c

The solution is b=-6, c=8

So the expression for the parabola is y=x2-6x+8

2. The axis of symmetry x=-2a b=3, l is the straight line x=31) when p is on the left side of the axis of symmetry, the distance from p to the axis of symmetry is 3-m, so the distance to the symmetry point e is 2 (3-m) = 6-2m

The distance from o to a is 4, and the height of the quadrilateral opaf is n

So there is. The area of the quadrilateral NAPAF = (6-2m+4)*n 2=20 and p is on the curve of the throwing hole, so there is m2-6m+8=n from the two equations to obtain m=......n=……

2) When p is to the left of the axis of symmetry, the distance from p to the axis of symmetry is m 3, so the distance to the point of symmetry e is 2 (m 3) = 2m 6

The distance from o to a is 4, and the height of the quadrilateral opaf is n

The area of the quadrilateral opaf = (2m 6+4)*n 2=20 and p is on the parabola next to the argument, so there is m2-6m+8=n from the two equations to obtain m=......n=……