Find the answers to two math questions steps .

, general score, (ab+bc+ac) abc=1

So, (ab+bc+ac)=abc

Shift, ab+bc=abc-ac, b(a+c)=ac(b-1)a+b+c=1

b(a+c)=ac(b-1)=b(1-b), shift.

ac+b)(b-1)=0, the solution is b=1 or ac=-b, when ac=-b, a+b+c=1, ac=a+c-1 is shifted and the same kind is combined: (a-1)(c-1)=0 is solved to get a=1 or c=1, so at least one of a, b, and c is equal to 12From x+1 y=y+1 z, x-y=1 z-1 y=(y-z) zy, and zy=(y-z) (x-y).

Similarly, zx=(z-x) (y-z), xy=(y-x) (x-z) Therefore, x 2*y 2*z 2=xy*yz*zx=(y-z) (x-y)*(z-x) (y-z)*(y-x) (x-z)=1

Question 2: Halfway through:

Let x+1 y=y+1 z=z+1 x=ss 3=(x+1 y)(y+1 z)(z+1 x)=3s+xyz+1 xyz

s^3-3s=xyz+1/xyz

s 3-3s) 2=(xyz) 2+1 (xyz) 2+2 if (xyz) 2=1, then.

s^3-3s)^2-4=0

s should be equal to plus or minus 1 or plus or minus 2

Now the problem translates to proving that s is equal to plus or minus 1 or plus or minus 2......

1 a+1 b+1 c (ab+ac+bc) abca+b+c=1 gives 1-a=b+c

1 a+1 b+1 c=1 gives ab+bc+ac=abcab+ac=abc-bc

a(b+c)=bc(a-1)

Substituting the top equation gives a(b+c)=bc(-(b+c))a+bc)(b+c)=0

When a+bc=0, a=-bc, so 1+bc=b+cbc-b=c-1

b(c-1)=c-1

b-1)(c-1)=0

So one of b and c is 1

When b+c 0 then b=-c, so 1 a=0 a=1 so there must be one of them

Because a=b=c=1 a=1 b=1 c, so a=1 a, because one is equal to his reciprocal, so 1 a=a 1, so a=1 is the same.

It's like the ...... in a practice paper that the teacher sent us before

Dude, you're not Huafu, are you?

1.You can get the right one:

a+b+c=a+b+c of abc

Multiply both sides of the equation by abc at the same time to get:

abc（a+b+c）=a+b+c

So A+B+C can be rescheduled.

Get: abc = 1

Now you think that a multiplied by b multiplied by c equals 1, then, if two numbers that are reciprocal to each other multiply equal to 1, in order to make their product 1, they can only multiply by one more 1, so that you can get that at least one of a, b, and c is 1

Oh, that's it, my language is not very good, whether I can understand it or not depends on your understanding.

Question 2: I want to send it to you again. It seems to be a little hard.

Upstairs (Crazy Sledgehammer) has to plagiarize the phenomenon of immorality.

a+b+c=1 yields 1-a=b+c

1 a+1 b+1 c=1 gives ab+bc+ac=abcab+ac=abc-bc

a(b+c)=bc(a-1)

Substituting the top equation gives a(b+c)=bc(-(b+c))a+bc)(b+c)=0

When a+bc=0, a=-bc, so 1+bc=b+cbc-b=c-1

b(c-1)=c-1

b-1)(c-1)=0

So one of b and c is 1

When b+c 0 b=-c, so 1 a=0 a=1 so there must be one for 1 and the second pass is being done

You don't have to pick it out, you don't have to take the test in such a question, and it's not very useful.

It's so hard in junior high school.

Alas, I sympathize with you.

You should ask me when I was in junior high school.

Today's children are smart and know how to make the most of resources, but your title is not very clear.

If you can't answer it, you can't answer it, isn't the question very clear? Anyway, I couldn't figure it out after thinking about it for a long time.

1 Set the point p coordinate to (x,0).

pa|²=(x-2)²+0-2)²=x²-4x+8|pb|²=(x+1)²+0+2)²=x²+2x+5|ab|²=(2+1)²+2+2)²=25①∠p=90°

then |pa|²+pb|²=|ab|²

x²-4x+8+x²+2x+5=25

x²-x-6=0

x1=-2x2=3

The point p coordinates are (-2,0) or (3,0).

a=90°|pa|²+ab|²=|pb|²

x²-4x+8+25=x²+2x+5

6x=28x=14/3

The coordinates of the point p are (14 3,0).

b = 90° then |pb|²+ab|²=|pa|²

x²+2x+5+25=x²-4x+8

6x=-22

x=-11/3

The coordinates of the point p are (-11 3,0).

To sum up: the point p coordinates that make the PAB a right triangle can be (-2,0) or (3,0) or (14 3,0) or (-11 3,0)2 There are two types of isosceles triangles.

1° when OP=OA=4

Let p(x,6-x),x 2 + 6-x)2 = 16

x 2 -6x + 10 = 0 x has no solution, so this triangle does not hold.

2° when Pa = Po

So p is on the straight line x=2.

The intersection of x=2 and x+y=6 is the coordinate of p, which is (2,4) so the height of poa is 4, so the area of poa s = 4 4 2=8

1.(2,0) (1,0) (2,0).

2.Did you make a mistake in the question?

2 set p(x,6-x).

a（4,0） o（0,0）

ap=√（4-x)²+6-x)²

op=√x²+（6-x)²

ao=4ap=ao

Get x = 5 7

x=5+√7 6-x=1-√7＜0

It doesn't fit the topic and leaves.

x=5-√7 6-x=1+√7

p(5-√7,1+√7)

ap=opx=2 6-x=4

It is the root of the original equation that fits the topic.

p(2，4)

op=ao 0 has no real root.

p(5-√7,1+√7)

s=1/2ah

p(2，4)

s=1/2ah=8

a-b）^2

a^2+b^2-2ab

a^2+b^2+2ab)-4ab

a+b）^2-4ab

So a-b = 32 = 4 on the side of the sail 2

x+1/√x)^2

x+1/x+2

The orange stupidity is x+1 x= 8=2 2

1.(a-b)^2=a^2+b^2-2ab=(a+b)^2-4ab=100-68=32

a2.The second question, your title is right, and if the complaint is correct, it is to open the root number directly, so that you can call it concisely.

Positive and negative root number 6

1: a+b=10 (a+b) of the state of bi square = 100 and (a+b) squared = the square of a + 2ab+b and the square of the pin (a-b) = the square of a - 2ab + b squared (a-b) square = (a + b) squared - 4ab = 100-4ab = 100-4ab = 100-4 17 = 100-68 = 32 a-b = 32 = 4 2

Sorry, I'll just ask this question.

1.(a-b) 2=(a+b) 2-4ab=100-68=32 a b, a-b=-4 root number 2

Xunhuai x+1 Blind withering x= Mochang Cave x+1 x= 6

1. This is an equal ratio series: the common ratio is 1+x, and the rest should be solved, right?

a0=1a1=1*(1+x)

a2＝1*(1+x)*（1+x）=1+x+x(1+x)a3＝1*(1+x)*（1+x）*（1+x）=(1+x+x(1+x))*1+x）＝1+x+x(1+x)+x(1+x)²

2. Take a look

m(m-n)-n(n-m)=m-square-mn-n-square+mn=m-square-n-square (m+n)(m-n)=12

m and n must be less than 12, how many should you think they should be equal to? 4，2

1.If x+y+z=30, 3x+y-z=50, x, y, z are all non-negative numbers, then the value range of m=5x+4y+2z is ( ).

Greater than or equal to 100, less than 110 greater than or equal to 110, less than 120, greater than or equal to 120, less than 130 greater than or equal to 130, less than 140 choose dx+y+z=30 (1), 3x+y-z=50(2)(1) *3+(2) to get 6x+4y+2z=140 more than m, the minimum is 0, and the maximum value is 140

2.In the plane Cartesian coordinate system, the coordinates of p1 are known to be (1,0), rotate it around the origin 30 degrees in the counterclockwise direction to get the point p2, extend op2 to p3 so that op3 = 2*op2, and then rotate the point p3 around the origin 30 degrees in the counterclockwise direction to get p4, extend op4 to the point p5 so that op5 = 2*op4, and so on, then the coordinates of the point p2010 are

2010*2*30 degrees = 120600 degrees.

i.e. p2010 on the x-axis.

op1=op2=1=2^0

op3=op4=2=2^1

op2010=2^1004

The coordinates of p2010 are (2 1004,0).

(1) Solution 1+x+x(1+x)+x(1+x) +x(1+x)2009 power.

The first and second terms are treated as one term (1+x), then the common factor (1+x) is extracted from the third term, combined into (1+x), and so on, and finally solved to the power of (1+x) 2008.

2) Solution: A negative sign is added to the second parentheses, and the original formula becomes m(m-n)+n(m-n)=12

m+n)(m-n)=12

Because m and n are positive integers, there are only two possibilities (m+n)(m-n)=6x2 or (m+n)(m-n)=4x3, and the second one does not meet the meaning of the topic and is rounded, so only (m+n)(m-n)=6x2, and solving this equation yields m=4 n=2

1. Finally, there are conditions for the formation of this figure, that is, AB be AE can form a triangle on three sides, and then it's up to you;

2. Rotate APC clockwise around point A so that the AC edge coincides with the AB edge to form AQB.

It is easy to know that pac = qab (that is, turn over), then qap=60 degrees, and aq=ap=10, then aqp is a regular triangle, then pq=10. And because qb=pc=6, bp=8, qp=10, qbp is a right angle. So qba+ pba=90 degrees.

Because QBa= PCA, PCA+ PBA=90 degrees. According to the outer angle relation, BPC= ABP+ ACP+ BAC, it's up to you.

Question 6, according to the definition of the simplest fraction, should be correct, I am very confused by your fork.

Question 7, xy=80000 can be obtained from the question

Substituting x=5 into the above equation yields y 15952

Question 8: If one side of the wall is x meters and the other side is y meters, then.

1° x 8.

xy=12x+2y≤

There are 2 integer enclosures: 6 meters long and 2 meters wide; It is 4 meters long and 3 meters wide, and the long side is the side that uses the wall.

2° x 8 hours.

xy=122x+2y≤

At this point, there is no solution to the group of inequalities.

PS: Question 6 should be correct.

Because 3-2x -1, then x<2

x-a 0, then x>a

The range of x is a, because x has four integer solutions, which are , -1, and -2

So -3c (taking into account a≠1).