Mathematics problems in the second year of junior high school are counter proof, and mathematics in

Suppose x, y, and z are all less than or equal to zero.

then x+y+z=a2-bc+b2-ca+c2-ab is less than or equal to zero.

and 2 (x+y+z).

c2-2bc+b2+b2-2ab+a2+a2-2ca+c2(c-b)2+(b-a)2+(a-c)2

is greater than or equal to zero.

In this case, x+y+z can only be equal to 0

c-b)2+(b-a)2+(a-c)2=0, i.e. a=b=c

This contradicts a, b, and c which are not exactly equal real numbers.

The null hypothesis does not hold up, so it ......

Assuming that xyz is not greater than zero, then x+y+z=a2-bc+b2-ca+c2-ab= is less than or equal to 0, and this equation is true only when abc is equal and equal to zero, which contradicts the condition. So if the assumption is not true, at least one of xyz is greater than zero.

Suppose that x, y, and z are all less than zero.

Then x+y+z=a2-bc+b2-ca+c2-ab2(x+y+z)=2a2+2b2+2c2-2ab-2ac-2bc2(x+y+z)=(a-b)2+(a-c)2+(b-c)2 know that it becomes a perfect equal, right?

Because the square is not negative and a, b, and c are not all equal real numbers, 2(x+y+z)=(a-b)2+(a-c)2+(b-c)2 must be greater than 0

Therefore, if the assumption is not true, then at least one of the propositions greater than zero in x, y, and z can be proven.

When we use the method of counterargument to solve a mathematical problem, we usually assume that the statement to be proved is false, and then derive contradictory conclusions through reasoning, thus proving that the original statement is true. Here's an example of proof using counter-proof:

Suppose we want to prove the statement that there is not a single student who cannot solve equations in the second semester of junior high school in the mathematics course.

We can use counter-evidence to prove that this statement is false. Suppose there is a student who does not know how to solve equations. Based on this assumption, we can draw the following conclusions:

This student was unable to answer an equation problem in math class.

Since this student can't solve equations, he may get the wrong answer on an exam or assignment.

This contradicts our original assumption that there is no one student who will not be able to solve an equation.

Therefore, we proved that there is no statement that any student in the mathematics course in the second semester of junior high school cannot solve equations is correct by the method of counterproof.

Note that this is just an example, and the actual math problem may require a more complex reasoning and proof process. The counterproof method is one of the commonly used proof methods in mathematics and can be used to solve various problems.

Suppose pb>=pc

Then the angle BCP > = angle CBP

Because ab=ac

So the angle acb angle abc

So angular acp< = angular abp

And because of the angle apb> the angle apc

So the horn bap "horn cap."

Because ab=ac

ap is a common edge.

So it can be obtained from the cosine theorem.

pc>pb

Doesn't match assumptions.

The conditions should state that L1 and L2 do not coincide, and the anti-explicit antecedent method is as follows

It is assumed that L1 is not parallel to L2 and that L1 and L2 do not coincide with the reincarnation.

L1 ABL2 is not perpendicular to AB and contradicts the known condition L2 AB.

The hypothesis is not true, i.e., L1 is parallel to L2

Proof: straight line a is not parallel to plane c; If the straight line A does not belong to C, then the relationship between the straight line A and the plane C is intersecting. However, the straight line B belongs to C, so the straight line A cannot be parallel to the straight line B, which contradicts the parallel between A and B in the title.

Therefore, if the assumption is not true, then a parallel plane c, can be proved.

Prove that assuming that A is not parallel to the plane C, and A is not on the plane.

Therefore, A and plane C must have an intersection point 0, and the point 0 is a straight line d parallel to B, then D is on plane C, so it does not coincide with A.

A is parallel to B.

Therefore, there are two non-coincident straight lines a, d and b are parallel to the point o, and the contradiction is proven.

Proof: Suppose A is not parallel to plane C

and a is not in c.

So because A intersects the plane C.

Because A and B are parallel.

Let the planes determined by a and b be

Then the line B is the intersection of the plane and the plane C, so the intersection of the line A and the plane C must be on the line B, which is parallel to A and B.

So the straight line a parallel plane c

Proof that two non-diameter strings ab and cd are bisected with each other, then abcd is a parallelogram a= c, b= d

a, b, c, d are all on the circle, a+ c=180 degrees, b+ d=180 degrees, so a= b= c= d=90 degrees, knowing that abcd is a rectangle, obviously ab and cd are the diameter of the circle, which contradicts the title, that is, it cannot be divided equally.

Suppose that the non-diameter two strings are bisected from each other.

Then these two strings must be over the center of the circle.

Because the string that passes through the center of the circle must be the diameter.

This contradicts the assumptions.

So the assumption is incorrect.

i.e. two non-diameter strings of a circle; They can't be divided equally with each other.

（1）。Assuming pb=pc, then abp is congruent with acp, then there is apb= apc, which contradicts the known, so the assumption is wrong and the problem is proven.

2) Assuming that a, b, and c are not divisible by 3, then the remainder of their division by 3 is only 1 and 2

If the remainder of c is the same as one of a and b, (or both) you might as well let the remainder of c and b be the same.

Since a 2 = c 2-b 2 = (c + b) (c-b), and c - b is a multiple of 3, there is a factor 3 in a, which means a is divisible by 3 which contradicts the hypothesis.

If the remainder of C divided by 3 and the remainder of A and B divided by 3 are not the same, then the remainder of A and B divided by 3 must be the same:

When the remainder of a, b divided by 3 is 1, the remainder of c divided by 3 is 2; A 2, and the remainder of b 2 divided by 3 is also 1, and the remainder of a 2 + b 2 divided by 3 is 2; The remainder of c 2 divided by 3 is 1. Thus A2+B2≠C2. This contradicts what is known.

In the same way, it is not possible when A and B are divided by 3 and the remainder is 2.

Thus, the original proposition is proven.