A few math problems for junior high school, urgent, urgent A few math problems for junior high schoo

Is there a problem with the first question?,It's impossible to be a parallelogram like this.。

The second problem is that the diagonal intersection is the origin, then the coordinates of point C and point D are (3,-1)(-7,-5).

then ab = 2 29 bc = 2 13

Perimeter = 4 29 + 4 13

1. A+c=b+7 is obtained from known conditions

The equation can be converted to b(a+c+1)+c2+16=0b(8+b)+c2+16=0

8b+b^2+c^2+16=0

b+4)^2+c^2=0

Because (b+4) 2>=0, c 2>=0

So dismantle the stupidity b = -4, c = 0, a = 3

Get b a=-4 3

294 967 295 (write 3 as (2 2-1) first, and then use the formula of square difference for continuous travel branches).

Ab+BC+B+C 2+16=0 is known

It is sorted out that Li Yu b(a+c+1)+c 2+16=0, and because a+c=b+7 is substituted into the above formula.

b(8+b)+c^2+16=0

8b+b^2+c^2+16=0

b+4)^2+c^2=0

Therefore, there is a lack of stalker (b+4) 2=0 and c 2=0

So b=-4, c=0, a=3

Get b a=-4 3

1) Replace c=7+b-a with ab+b*(7+b-a)+b+(7+b-a) 2+16=0

Simplification: 8b+b 2+16+(7+b-a) 2=0b+4) 2+(7+b-a) 2=0

So b+4=0 7+b-a=0 b=-4, a=3 and then we get 2 32-1=

(1) Treating x +y as an unknown number can be solved to get x +y = 4 (2) original formula = (a + b) (c + d) = 1997 Because 1997 is a prime number, so a + b = 1 or 1997, c + d = 1997 or 1, then a + b + c + d = 1998

3）2001^3-2x2001^2-1999=2001^2x(2001-2)-1999=1999x(2001^2-1)

So: original = 1999 2002

(1) Let x +y = a a*(a-1) = 12 (a-1 2) = 49 4

a-1/2=±7/2

a=1/2±7/2

a = 4 or -3

So x +y = 4 or x + y = -3

Since x +y 0 so x +y = 4

(1)x²+y²=4

3) Defactoring = 2001 (2001-2)-1999 2001 (2001+1)-2002=1999(2001-1) 2002(2001-1)=1999-2002

The second title is 1998

The first question can be given as an unknown number x +y. The final answer is 3 or 4

1.Let x +y = t, the original formula can be reduced to t -t-12 = 0

The solution yields t=4 or t=-3

The following is thinking about it for a while.

estimation method).

2.(1) < (both sides squared at the same time).

2) > (both sides squared at the same time).

3) > (estimate: three changes number 9 2...., minus one is about 1....4) > (estimate: root number 3, minus 2

and minus [(root number 2) 3].

5) < (first add 1 on both sides at the same time, (root number 5) 2 and 1 6 comparison, and then square comparison at the same time).