A few math problems in the second year of junior high school. Everybody help to untie the pull down

Let the number written by Xiao Ming be x and the number written by Xiao Liang be y, then it will be derived from the title.

x+y=148,(1)

x-y=2,(2)

1) + (2) get.

2x=150, so x=75, so y=73, so Xiao Ming. The numbers written by Xiaoliang are 75 and 73 respectively

2.Let the first three numbers are x, the fourth number is y, then the fifth, the sixth number are y+1, y+2 respectively, from the meaning of the title 3x+y+y+1+y+2=10(y+1)+y+2, so 3x+3y+3=11y+12, so 3x=8y+9, when y=0, x=3, meet the topic;

When y=1, x=23 3, not an integer from 0 to 9, round off;

When y=2, x=34 3, not an integer from 0 to 9, rounded off;

When y=3, x=45 3=15, beyond the range of integers from 0 to 9, rounded;

So x=3, y=1, so this six-digit number is 333012

1.Solution: Let the number written by Xiao Ming be x, and the number written by Xiao Liang be y, then.

x+y＝148

x-y 2 solves:

x＝75，y＝73

2.Solution: Let the three numbers on the left be x, and the three numbers on the right are y, y+1, and y+2 respectively

So. x+x+x+y+y+1+y+2 (y+1) 10+y+2. 3x+3y+3＝11y+12

3x＝8y+9

From the title, x and y are both whole numbers, so the six-digit number x 3 and y 0 are 333012.

Solution: (1) Let the number written by Xiao Ming be x and the number written by Xiao Liang be y

then there is x-y=2

x+y=148

x=75 y=73

2) Let the first digit of this six-digit number be x, and the fourth digit is y, then there is 3x+y+y+1+y+2=10(y+1)+y+23x+3y+3=11y+12

3x-8y=9

3x=8y+9

And because x is less than 10 positive integer y is an integer less than 10, so 3x is less than 30, that is, 8y+9 is less than 30, you can know that y can take 0, 1, 2, and because when y takes 1, 2 is x, not an integer, so you know y=0 x=3, so these six digits are 333012

Let Xiao Ming and Xiao Liang write x and y respectively

x-y=2x+y=148

Add the two formulas to 2x=150

x=75 So Xiao Ming and Xiao Liang write 75 and 73 respectively, let this number from left to right are a, a, a, b, b+1, b+23a+3b+3=10(b+1)+b+2

3a-8b=9

So a=3, b=0

So this six-digit number is 333012

By: Anonymous 2-22 17:20

Let the number written by Xiao Ming be x and the number written by Xiao Liang be y, then it will be derived from the title.

x+y=148,(1)

x-y=2,(2)

1) + (2) get.

2x=150, so x=75, so y=73, so Xiao Ming. The numbers written by Xiaoliang are 75 and 73 respectively

2.Let the first three numbers are x, the fourth number is y, then the fifth, the sixth number are y+1, y+2 respectively, from the meaning of the title 3x+y+y+1+y+2=10(y+1)+y+2, so 3x+3y+3=11y+12, so 3x=8y+9, when y=0, x=3, meet the topic;

When y=1, x=23 3, not an integer from 0 to 9, round off;

When y=2, x=34 3, not an integer from 0 to 9, rounded off;

When y=3, x=45 3=15, beyond the range of integers from 0 to 9, rounded;

So x=3, y=1, so this six-digit number is 333012

Anonymous 2-22 17:22

Let Xiao Ming write x, and Xiao Liang write y

So x+y=128

x-y=2

Let Xiao Ming write x, and Xiao Liang write y

So x+y=128

x-y=2, so x=75 y=73

Because 4x-3 is a factor of the original formula, it must be divisible by dividing the original formula by 4x-3, and the remainder of the tan pi is a+6, because the rent is divisible, so a+6=0, a=-6).

2.The value is 1. Because 3y 2-2y+6=8, so 3y 2-2y=2, so 3xy 2-2xy=2x gets 3 2y 2-y=1, and substitutes it into the original formula to get the original formula =1

3.The original formula = m 2 + 2m + 1 + n 2-6n + 9 = (m + 1) 2 + (n - 3) 2 = 0, because the sum of two non-negative numbers is equal to zero, so these two non-negative numbers are zero, i.e. m+1=0, n-3=0, m=-1, n=3, mn=-3

4.Because x+y-2=0, x+y=2Original = (x+y) 2(x-y) 2-8x 2-8y 2=4(x-y) 2-8x 2-8y 2 gets the original formula =-(4x 2+8xy+4y 2)=-2x+2y) 2=-[2(x+y)] 2=-16

5.The original formula = (2x+3y)(2x-3y)=31, because 31 is a prime number, and x,y are positive numbers, so 2x+3y=1,2x-3y=31 or 2x+3y=31,2x-3y=1, solve the system of equations to get x=8,y=-5 or let the difference x=8,y=5, because x,y are positive numbers, so the first case is rounded, so the final result is x=8,y=5

1/a+1/b=4/a+b (a+b)/ab=4/a+b (a+b)^2-4ab=0 a^2+b2=2ab

b/a+a/b=b^2+a^2/ab=2ab/ab=21/x-3+7=x-4/3-x

2 1 (x-3)+7=(x-4 3-x) multiply both sides by x-3 to get (x≠3).

1+7(x-3)=-x+4

8x=24x=3

The original equation has no solution and has an additional root x=3

3 Is the third question correct?

4 a+2/x+1=1

x+1=a+2

x=a+1<0

a<-1

1 a+1 b=4 a+b left pass is divided into (a+b) ab=4 (a+b).

Simplifying (a+b) 2 ab=4, i.e., there is (a 2+b 2) ab+2=(b a+a b)+2=4, so b a+a b 4-2 2

1 (x-3)+7=(x-4) (3-x) The root of the fractional equation is that the denominator is 0

So x-3=0

x=3 Original formula = [(m-2n) 2-3n 2] (m-2n)(m+2n) is not simplified, is there a problem with the problem?

a+2/x+1=1

a(x+1)+2=x+1

ax+a+2-x=1

ax-x=-a-1

x=-(a+1)/(a-1)

Because the solution is a non-positive number.

So -(a+1) (a-1)<=0

a+1)/(a-1)>=0

Step-by-step discussion: a+1>=0 and a-1>0

i.e. a -1 and a > 1

So a>1

a+1<=0 ,a-1<0

a<=-1 ,a<1

So a>1

Based on the above, it can be seen.

A>1 or A-1

1. From the condition that it is easy to deduce that square A plus square B is equal to 2AB, then the original formula is equal to 2

1.When a=2, b=one-seventeenth, the minimum value of p is equal to 1987, and b, c, and d are all equal to zero, so ab+cd=0

3.< 1>x squared + y squared - 2x + 12y + 40 = x squared - 2x + 1-1 y squared + 12y + 36-36 + 40 = (x-1) squared + (y + 6) squared + 4 (so no matter x, y is a positive number).

I'm just watching and taking points.

square + 17b square - 16a - 34b + 2004 = 2 (a square - 8a + 16) + 17 (b square - 2b + 1) + 1955 = 2 (a - 4) square + 17 (b - 1) square + 1955 so when a = 4 b = 1 value minimum = 1955

1> x square + y square - 2x + 12y + 40 = (x square - 2x + 1) + (y square + 12y + 36) + 3 = (x-1) square + (y + 6) square + 3

Therefore, no matter how many xy is, the number obtained is 3, so it is a positive number.

Now that I have the answers to the first four questions, I'll answer the fifth question.

x 2-yz=z 2-xy, you can get x 2-z 2=yz-xy, that is, (x-z)(x+z)=-y(x-z), in two cases, 1, x-z=0, the equation holds, that is, x=z, y 2=xz=x 2, so y=x, substituting the sub-verification of the formula, it can be obtained only when x=y=z=0, it is true;

2. When x-z≠0, x+z=-y, that is, x+y+z=0