Five junior high school Olympiad questions, five fifth grade Olympiad questions

, b-c=-3, c-d=5, find (a-c)(b-d) (a-d).

Solution: a-c=a-b+b-c=2-3=-1

b-d=b-c+c-d=-3+5=2

a-d=a-b+b-c+c-d=2-3+5=4

So (a-c)(b-d) (a-d)=-1*2 4=-1 2

, ax 3-bx+1=-17, and when x=-1, find the value of 12ax-3bx 3-5.

Solution: ax 3-bx+1=-17 when x=2

8a-2b+1=-17

4a-b=-9

x = -1, 12ax-3bx 3-5

12a+3b-5

3(4a-b)-5=22

Find (3a-7ab-6b) (-2b+3ab+a).

Solution: 1 b-2 a=2

2ab=a-2b

3a-7ab-6b）/（-2b+3ab+a）

6ab-7ab）/（2ab+3ab）

Find x+y+z

Solution: x (a-b) = y (b-c) = z (c-a) = k

then x=ak-bk, y=bk-ck, z=ck-ak

x+y+z=0

Find a 0+a 2+a 4+a 6+a 8+a 10

Solve x=1 instead of a 0 x 10 + a 1x 9 + a 2 x 8+....+a 9x+a 10=(x 2-x+1) 5.

aˇ0+aˇ1+aˇ2+…+aˇ9+aˇ10=1^5=1

Replace x=-1 with a 0 x 10+a 1x 9+a 2x 8+....+a 9x+a 10=(x 2-x+1) 5.

aˇ0-aˇ1+aˇ2+…-aˇ9+aˇ10=3^5=243

Add the two formulas to give 2 (a 0 + a 2 + a 4 + a 6 + a 8 + a 10) = 244

So a 0 + a 2 + a 4 + a 6 + a 8 + a 10 = 122

1、(a-b)+（b-c）=a-c=-1 （b-c）+（c-d）=b-d=2

a-b）+（b-c）+（c-d）=a-d=4

a-c）（b-d）÷（a-d）=-1/2

2. Bring x in, then we know 8a-2b+1=-17, and find -12a+3b-5

12a+3b-5=-3/2(8a-2b+1)-7/2=22

3. Divide the semicolon up and down by ab at the same time, and then substitute 1 b-2 a=2.

4. Let x (a-b) = y (b-c) = z (c-a) = k

then x=ka-kb, y=kb-kc, z=kc-ka

x+y+z=0

5. Let x=1, then a 0+a 1+a 2+....+a 9+a 10=(1-1+1) 5=1 (1).

Let x=-1, then a 0-a 1+a 2+....-a 9+a 10=(-1+1+1) 5=1 (2).

1)+(2)2*(a 0+a 2+a 4+a 6+a 8+a 10)=2

i.e. a 0+a 2+a 4+a 6+a 8+a 10 =1

b-d=(b-c)+(c-d)=2

a-d=(a-b)+(b-c)+(c-d)=4, so (a-c)(b-d) (a-d)=

2.Bringing x=2 into ax 3-bx+1=-17 gives you 8a-2b=-18 and you get 4a-b=-9

When x=-1, 12ax-3bx 3-5=-12x+3b-5=-3*(4a-b)-5=22

(a-2b) ab=2

So a-2b=2ab

3a-7ab-6b=3(a-2b)-7ab=-ab2b+3ab+a=-(a-2b)+3ab=ab, so (3a-7ab-6b) (-2b+3ab+a)=-1

a-b=2 ①

b-c=-3 ②

c-d=5 ③

=a-c=-1 ④

=b-d=2 ⑤

+=a-d=4 ⑥

a-c）（b-d）÷（a-d）=

When x = 2, ax 3-bx+1=8a-2b+1=2*(4a-b)+1=-17

4a-b=-9

When x=-1, 12ax-3bx 3-5=-12a+3b-5=-3*(4a-b)-5=22

1/b-2/a=2

a-2b=2ab

3a-7ab-6b）/（-2b+3ab+a）=[3a-6b-7ab]/[a-2b+3ab]

6ab-7ab)/(2ab+3ab)=-1/5

x/（a-b）=y/（b-c）=z/（c-a)

y=（b-c)x/（a-b）

z=(c-a)x/（a-b)

x+y+z =x+（b-c)x/（a-b）+(c-a)x/（a-b)=0

Let x=1a 0x 10+a 1x 9+a 2x 8+....+aˇ9x+aˇ10=（x^2-x+1）^5

can be turned into. aˇ0+aˇ1+aˇ2+..aˇ8+aˇ9+aˇ10 =1 ①

Let x=1a 0x 10+a 1x 9+a 2x 8+....+aˇ9x+aˇ10=（x^2-x+1）^5

can be turned into. aˇ0-aˇ1+aˇ2+..aˇ8-aˇ9+aˇ10 =3^5 ②

aˇ0+aˇ2+aˇ4+aˇ6+aˇ8+aˇ10 =

1, 1)+2) gets: a-b b-c 2 3, a-c = -1 The same goes for b-d 2

The sum of the three formulas yields: a-d 2 3 5 4

a-c）（b-d）÷（a-d）＝－1/2

2, x = 2, ax 3-bx+1 = -17

8a-2b=-18

4a-b=-9

When x=-1, 12ax-3bx 3-5=-12a+3b-5=-3*(4a-b)-5=22

4. Let x (a-b) = y (b-c) = z (c-a) w then: x+y+z (a-b)w+(b-c)w+(c-a)w=0

From a-b=2 and b-c=-3, it is obtained: (a-b)+(b-c)=a-b+b-c=a-c=2-3=-1

From b-c=3-,c-d=5, it is obtained: (b-c)+(c-d)=b-c+c-d=b-d=-3-5=-8

From a-b=2, b-c=-3, c-d=5, it is obtained: (a-b)+(b-c)+(c-d)=a-b+b-c+c-d=a-d=2-3+5=4

Therefore: (a-c)(b-d) (a-d)=-1*(-8) 4=2 There is no time to do the following questions, hehe.

5*2+4*2=18

18 (2-1)=18 male: 18+4=22, female: 22+5*2=32

39+9=18 B18 4-1=6 A 6*4=24450*2=100 50*8+60*5-100=600 60-50=10 600 10=60 60+8=68 Ming's family left school? m 50 * 68 = 3400

520 + 25 = 45 45 * 10 = 450 450 The two towns are far apart.

248 6=8 48 4=12 Area=12*8=96

1. From the first condition: your classmates have 10 more people than their male classmates, and the second condition: if the condition is true, there are 18 (10+4+4) more male classmates than male classmates, and the number of female classmates is now twice as many as male classmates, so there are now 18 male classmates and 36 female classmates.

It was concluded that there were 22 male students and 32 female students.

2. You can draw a picture on the paper, so that the length and width of the rectangle are extended to the value in the question, at this time, the increased area is divided into three pieces, one is a rectangle of 6 * 4 = 24 square meters, and the remaining two are 4 * long, 6 * wide rectangles, so the remaining two pieces have a total of 24 square meters, if the length and width must be integers, then the length is 3, and the width is 2 for the positive solution (to meet the length than the width of the length, but if it can not be an integer, then there are infinite solutions), so the original area is 6

3. The unit price of A is four times that of B, then the unit price of A is 3 times more than the unit price of B, and the second condition shows that the unit price of A is 18 yuan more than B, so the unit price of B is 18 3 = 6, and the unit price of A is 4 * 6 = 24

4. Xiao Ming's time to go out every time is the same, and the time from class is certain, so assuming that Xiao Ming has been walking for the last five minutes of the second bell plan, he will walk 60 * 5 + 50 * 8 = 700 meters more than the first plan during class, and the speed of the second plan is 10 meters per second faster than the first one, 700 10 = 70 minutes, which means that Xiao Ming walked for 70 minutes at a speed of 60, so it took a total of 72-5 = 67 minutes to get to school, so the distance from school to home is 65 * 60 + 50* 2 = 4000 meters.

5. After C meets B, he and A go in the opposite direction, so the distance between the two of them is 10*(25+20)=450, and this distance is also the distance between B and A, and it took 450 (minutes) during the period, which means that C met A at the 180th hour, so the distance between the two towns is 180*(25+m.

1. We know that the number of male and female students is the same, so if the number of female students participating is reduced by 5, the same.

If the number of male students is reduced by 4 and the number of female students is increased by 4, then the difference between the number of male and female students will increase to 5 + 5 + 4 + 4 = 18, because the number of female students is twice as many as the male students at this time, so 18 people are the number of male students at this time, so 22 male students and 32 female students at this time.

2, 48 divided by 6 = 8, so 8 meters wide, 48 divided by 4 = 12, so 12 meters long, the original area of 96 meters, these questions are almost the same, but I think it is a very normal problem, I don't know how it became an Olympiad problem?

3. B: 9 2 (4-1) = 6 (yuan) A: 6 4 = 24 (yuan).

2 = 100 (m) 60 5 = 300 (m) 300 50 = 6 (min) 60 6 + 100 = 460 (m).

by x1x2x3....x2007=x1-x2x3…x2007=x1x2-x3…x2007=…=x1x2x3...x2006-2007=1

x1x2x3...x2006-1/x1x2x3...x2006=1 then: x1x2x3....x2000-1/x1x2x3...x2000=1

x1x2x3...x1999-1/x1x2x3...x1999=1 to get x1x2x3....

x2000 = (1 plus or minus root number 5) divided by 2x1x2x3....x1999 = (1 plus or minus root number 5) divided by 2x2000 = x1x2x3....x2000/x1x2x3...

x1999 = 1 or (plus or minus root numbers 5-3) 2

It's actually quite ......

First of all, neither x1 nor x2007 equals 0, and x1x2 ......x2007=1, set an=x1x2x3x4......xn（n=1，2，3……Anderson et al., 2007), apparently an is not equal to a(n-1).

So an 2-1=an

Then find a2000 and a1999, and then divide them to find x2000

Known x1x2x3....x2007=x1-x2x3…x2007=x1x2-x3…x2007=…=x1x2x3...x2006-2007=1

x1x2x3...x2006-1/x1x2x3...x2006=1 yields: x1x2x3....x2000-1/x1x2x3...x2000=1

x1x2x3...x1999-1/x1x2x3...x1999=1 to get x1x2x3....

x2000 = (1 plus or minus root number 5) divided by 2x1x2x3....x1999 = (1 plus or minus root number 5) divided by 2x2000 = x1x2x3....x2000/x1x2x3...

x1999=1 or (plus or minus root numbers 5-3) 2 The result is out.