Seek the master of mathematics in the second year of junior high school, and ask for super difficult

Use the commutation method to set:

a=a^2,b=b^2,c=c^2

Original formula = a 2 + b 2 + c 2-2ab-2bc-2ac sub multiplied by 2 multiplied by 1 2 both:

1/2(a^2-2ab-b^2+b^2-2bc+c^2+a^2-2ac+c^2)

After the recipe: 1 2((a-b) 2+(b-c) 2+(a-c) 2) so greater than or equal to zero.

The split term is a4-a2b2-a2c2+ b4-b2c2-b2a2+ c4-c2a2-c2b2

Then a4-a2b2-a2c2=a2(a2-b2-c2)=a2(a-b+c)(a-b-c)<0

In the same way b4-b2c2-b2a2 "0 c4-c2a2-c2b2<0

So the result is negative.

Support! There are doubts about the answer to the second floor:

Original formula = a2+b2+c 2-2ab-2bc-2ac

Multiply by 2 and then multiply by 1 2 both:

1/2(a^2-2ab-b^2+b^2-2bc+c^2+a^2-2ac+c^2)

The brackets in parentheses are multiplied by 2 for a 2, and -2ab, -2bc, and -2ac seem to have forgotten to multiply by 2.

The quadrilateral abcd is trapezoidal, so ab cd. a=90°, d=90° (parallel theorem, complementary internal wrong angles) through point b to make a straight line bm intersection dc extension line at point m. It can be known that dm=ab=2, and cd=1, so cm=2-1=1.

It is easy to prove that the BCM is at right angles (m=90°) and cm=1 and bc=3. It can be obtained that BM = 2 times the root number under 2. The quadrilateral abmd is rectangular.

So ad=bm=2 times the root number under 2. e is the midpoint of AD, ae=ed=AD2=2 under the root number. From the proportional relation of the edges, we can prove abe dec, from which dec= eba eba+ abe=90° dec+ abe=90° ceb=90° ce be.

Solution: Let ae=ed=a, which is obtained by the Pythagorean theorem (1 2+a 2)+(a 2+2 2)=?=3^2...If true.

Simplified to a 2=?=2

From (2a) 2+1 2=3 2 gives a =2.So ce be

Parallel lines over e as AB cross BC to F

Because E is the AD midpoint.

then ef=(ab+cd) 2=

And because f is the midpoint of BC (provable).

BC2=so CEB=90

The quadrilateral amen and efcg are diamond-shaped.

Because the quadrilateral ABCD is rhomboid, mg parallel ad parallel bc, nf parallel ab parallel cd

So the quadrilateral bcgm, quadrilateral fcdn, quadrilateral amen, quadrilateral efcg are all parallelograms.

So BM=CG,ND=CF; i.e. cg=cf;

So the quadrilateral EFCG is a diamond.

And because mg=bc, nf=ab, ab=bc, en=nf-cg, em=mg-eg, em=en

So the quadrilateral amen is also a rhombus.

s△aef=s△abc=s△kcd

Let bc=a, ac=b, ab=c

Then, in AEF, AE=AC=B, AF=AB=C, then we get from the sinusoidal theorem, S AEF=(1 2)*AE*AF*SIN EAF

1/2)bc*sin∠eaf

1/2)bc*sin[360°-∠cae-∠baf-∠cab](1/2)bc*sin[360°-90°-90°-a](1/2)bc*sin(180°-a)

1/2)bcsina

s abcSo, when abc is a right triangle and ab=1 and bc=2, s aef=s kcd=s abc=(1 2)*1*2=1

a, x, y are not equal to each other, and a-y can be opened to know a-y>0, so a(y-a) can be opened to know a<0 or a=0;x-a can be opened to know x-a>0, so a(x-a) can be opened to know a>0 or a=0;So a=0 , so x=-y and not equal to 0, so (3x +xy-y) x -xy+y )=1 3

The upstairs statement is very correct, I don't need to add it, let's give him a share.