Urgent! Points! A math problem in the first year of junior high school, answer well and add 50

Because the addition of PA+PB>AB, PB+PC>BC, PA+PC>AC yields: 2 (PA+PB+PC)>AB+BC+AC=3, i.e., PA+PB+PC>3 2

Extend the bp to AC on D

Because ab+ad>ad, i.e., ab+ad>bp+pdpd+dc>pc

The addition of the two formulas yields: ab+ad+pd+dc>pb+pd+pc, so ab+ac>pb+pc

In the same way, BC+AC>BP+AP, AB+BC>AP+PC are added to the formula: 2(AB+BC+AC)>2(PA+PB+PC) so PA+PB+PC In summary: 3 2 PA+PB+PC 2

Or. In the equilateral triangle ABC, p is on the triangle, so the sum of the two terms in Pa+PB+PC equals 1

Then the height of the 3 2-1 triangle is 1 2

The shortest distance from C to AB is high, so PC is greater than 1 2PC<1 (in a triangle, the side of PC is 60° Ca or the side opposite to CB is greater than 60°, and the large side is greater than 60°).

So 1 2 is: 3 2 Pa + PB + PC 2

Proof: In the equilateral triangle ABC, p is on the triangle, so the sum of the two terms in Pa+PB+PC equals 1

Then the height of the 3 2-1 triangle is 1 2

The shortest distance from C to AB is high, so PC is greater than 1 2PC<1 (in a triangle, the side of PC is 60° Ca or the side opposite to CB is greater than 60°, and the large side is greater than 60°).

So 1 2 is: 3 2 Pa + PB + PC 2

1.-a²+4ab-b²=-(a²+b²-4ab)=-(a²-2ab+b²-2ab)=-(-10+16)=-6

This question should skillfully use the known conditions and construct what you want...

All the questions in this question revolve around p and q, merge similar items first, simplify what you seek, you can get twice the result with half the effort, and reduce errors.

p-[q-2p-(-p-q)]=2q=2(a²-3ab+b²)=2a²-6ab+2b²

3.Look at the title, you will find the common factor (3x-y), propose the common factor, get:

2-4+5-6) (3x-y)= -3(3x-y)Calculation(3x-y)= -1, Original = -3 (-1)=3 This is my own doing, although the question is not difficult, but the keyboard is very hard...

1) The original formula can be reduced to: -(a -4ab+b) because a -2ab=-10, b -2ab=16 so -(a -4ab+b )=-(a -2ab+b -2ab)=-(16-10)=-4

2) The original formula can be reduced to: 2p-2q (remove brackets) and substitute p=a +3ab+b and q=a -3ab+b to obtain 2 (a +3ab + b) -2 (a -3ab + b) that is, 2 (a +3ab + b -a +3ab-b ) = 2 (6ab) = 12ab; When a=1, b=-1, 12ab=12(-1*1)=-12

3) The original formula can be reduced to: (3x-y)(2-4+5-6)=-3(3x-y), substituting x=-5/12, y=-1/4, obtains, =-3(3x-y)=-3(minus five-quarters + quarters)=3

1.Solution: a -2ab = -10....Eq. 1b -2ab = 16....Equation 2, Equation 1, plus Equation 2:

a²-4ab+b²=26

So -a +4ab-b = -26

2 From the question, it can be obtained that p-[q-2p-(-p-q)] is equal to p-q+2p-p-q is equal to 2p-2q

Bring in p,q to get 2a +6ab + 2b -(2a -6ab + 2b ) to get 12ab

When a=1 and b=-1, the value of the equation is 12 times 1 times 1 times 1 to get 1232(3x-y)-4(3x-y)+5(3x-y)-6(3x-y)=6x-2y-12x+4y+15x-5y-18x+6y=-9x+3y

x = -5/12, y = -1/4

So the original formula is equal to -9(-5 12)+3(-1 4)=15 4-3 4=12 4=3

Do it yourself.

(1) Use the whole method: the original formula =-[(a 2-2ab)+(b 2-2ab)]=-[-10+16].

6 (2) Simplification: original formula = 2 (p-q).

12ab is then substituted: when a = 1, b = -1, the original formula = -12 (3) the original formula = 6x-2y-12x+4y+15x-5y-18x+6y=-9x+3y

Substituting x,y yields: original = 3

(1) You can convert the original equation to -(a -4ab+b), and add the first two equations to 6

2) P=Q+6AB can be obtained, and P=Q+6AB can be brought into P-[Q-2P-(-P-Q)] to obtain the original formula =12AB, when A=1, B=-1, the original formula =-12

3) Original formula = (2-4 + 5-6) (3x-y) = -3 (3x-y), when x = -5/12, y = -1/4, the original formula = 9/2

(1) -a +2ab=10 because a -2ab=-10 , -b +2ab=-16 because b -2ab=16, the two sides of the equal sign are added to -a +4ab-b =-6;

2)p-[q-2p-(-p-q)]=p-[q-2p+p+q)]=p-[2q-p]=2p-2q =2（p-q)

Substituting p=a +3ab+b and q=a -3ab+b into the equation yield:

2[a²+3ab+b²-(a²-3ab+b²)]=2[a²+3ab+b²-a²+3ab-b²]=2[6ab]=12ab

When a=1, b=-1, 12ab=12(-1*1)=-12The value of the equation is -12

3)2（3x-y）-4（3x-y）+5（3x-y）-6（3x-y）=(2-4+5-6)（3x-y）=-3（3x-y）=-9x+3y ;

where x = -5/12, y = -1/4 at -9(-5 12)+3(-1 4)=(15 4) +3 4)=12 4=3.

(1) a -2ab = -10, b -2ab = 16 so -a +2ab = 10, -b + 2ab = -16, the addition of the two formulas gives -a + 4ab-b = 10-16 = -6;

2）p-[q-2p-(-p-q)]=2p-2q=2（p-q）=6ab；a=1, b=-1, the result is -6

3）2（3x-y）-4（3x-y）+5（3x-y）-6（3x-y）=6x-2y-12x+4y+15x-5y-18x+6y=-8x+3y

Replace x = -5/12 and y = -1/4

Brought in: -49 12

(1) a -2ab=-10, -a +2ab=10, b -2ab=16, -a +2ab-(b -2ab)=-a +4ab-b =10-16=-6

2)p-[q-2p-(-p-q)]=p-(q-2p+p+q)=p-q+2p-p-q=2p-2q=2a²+6ab+b²-（2a²-6ab+2b²）=12ab=-12

3) 2 (3x-y) -4 (3x-y) + 5 (3x-y) -6 (3x-y) = (3x-y) (2-4 + 5-6)

3（3x-y）

Substituting x = -5/12 and y = -1/4 into -3 (-5/4 - 1/4) = -3 (-3/2) = 9/2

Solution: Let Xiao Zhang use at least x cubic meters of water this month.

10×2x＞30

x 15A: Xiao Zhang uses at least more than 15 cubic meters of water this month.

Not more than 10 cubic meters, the maximum water fee is.

Xiao Zhang's water bill for a month is not less than 25 yuan, more than 15, so Xiao Zhang uses at least x cubic meters of water this month if he uses tap water of more than 10 cubic meters.

15+2x-20≥25

2x-5≥25

x≥(25+5)/2=15

Xiao Zhang uses at least 15 cubic meters of water this month.

If x is less than 10 and greater than or equal to 25

or x greater than 10 10 times greater than or equal to 25

You can solve it.

If the water consumption is at least x cubic meters, there is: 10*

Solution:1

a-4b-[(a+b)÷2]

2a-8b-a-b)/2

a-9b)/2

a^2-3a+1-9(2a^2+a-4)]/2=(-17a^2-12a+37)/2

2a^3+5a^2+2a-1-(8a^3+3a^2+10a-4)=-6a^3+2a^2-8a+3

2a^3+5a^2+2a-1+8a^3+3a^2+10a-4=10a^3+8a^2+12a-5

The other polynomial is -6a 3+2a 2-8a+3 or 10a 3+8a 2+12a-5

c=-x^2+2x-3+2x^2-3x+1=x^2-x-2c=x^2-x-2

1 a=-1 substituting the two formulas a and b gives a=5, b=-3a-4b-[(a+b) 2]=16

2The difference between the 2 polynomial 2a +5a +2a-1 and another polynomial is equal to 8a +3a +10a-4

Then the required polynomial is equal to (2a +5a +2a-1) -8a +3a +10a-4) = -6a +2a -8a +3

3a+b+c=0, so c=-a-b=x -x-2 (a=x -2x+3, b= -2x +3x-1).

When a=-1.

1.Solution: a-4b-[(a+b) 2].

a-4b-a÷2+b÷2]

a-a/2+(-4b+b/2)

a-a/2-4b+b/2

When a=-1.

Original formula: a-a 2-4b+b 2==

2.Solution: 8a +3a +10a-4-(2a +5a +2a-1)=8a +3a +10a-4-2a -5a -2a+1=(8-2)a +(3-5)a +(10-2)a+(-4+1)=6a -2a +8a-3

The third question is not solved because I don't have time. Anyway, it's very simple, that question

1.When a=-1, a=5, b=-3, so the algebraic value is 16

2.The required polynomial is the difference between 2a +5a +2a-1 and 8a +3a +10a-4, and the result is -6a +2a -8a+3

3.c = -(a+b), the result is x -x-2

It's hard to see this expression.

1.According to the known conditions, knowing that the two terms are of the same kind, then we can get that the powers of a and y are the same.

The system of equations obtained is as follows:

m=n+1n=2

solution, m=3, n=2

Bringing in the expression requested below gives the result of 33 2

2.From the known, 3*3*3*3*3*a+3*3*3*b+3*c-5=11, then 3*3*3*3*a+3*3*3*b+3*c=16

Bringing in -3 yields -3*3*3*3*3*a-3*3*3*b-3*c-5=-16-5= -21

So the result is -21

According to the meaning of the question, in order for their sum to be a monomial, the two must be the same kind of terms before they can be combined into a single mimial, so the exponents of a are equal to the exponents of a, and the exponents of y are also equal.

So. n=2

m=n+1, so m=3, n=2

Therefore, the values of m and n are substituted into the requirement value expression respectively

Bring +3 and -3 into the expression respectively

3a) 5+(5b) 3+(3c)-5=11 can be reduced to: (3a) 5+(5b) 3+(3c)=16-3a) 5+(-5b) 3+(-3c)-5=?

It can be reduced to: [3a) 5+(5b) 3+(3c)]-5=?

So the end result is: -21