Junior 2 Mathematics to detail the process, thank you .

Subtract 4ab from both sides of the original formula, then the original left side = a 2*b 2+a 2+b 2-4ab+1=a 2*b 2-2ab+1+a 2-2ab+b 2=(ab-1) 2+(a-b) 2, and because the original right side = 0, so ab-1=0, a-b=0, then a=b, ab=1, so a=1, b=1, or a=-1, b=-1

It is known that the two roots of the equation x2+(2+a)x+1+a+b=0 are x1 and x2and 00,2+a<0

then a<-2, =>0 b>1

then b a+1+1 a<0 (due to a<-2).

1/a<-1-b/a

Let y=x-1

Then (y+1) +2+a)(y+1)+a+b+1=0 gives y +(4+a)y+2a+b+4=0

Due to x1<10

0<2+b a+4 a<2+b a+4(-1-b a) (1 a<-1-b a).

b/a<-2/3

2+b/a+4/a>0

b a>-2-4 a (because a<-2,-4 a>0) then b a>-2

In summary, b a (-2, -2, 3).

After traveling for 2 hours, Jiashan Lakuan walked 6 2 = 12 kilometers to make it bright.

B walked for 1 hour, then walked 5 1 = 5 km.

The walking route of A and B forms a straight angular triangle, which is to find the hypotenuse.

then hypotenuse = root number (12 +5) = 13 km.

A: 13 kilometers away.

13 kilometers of noisy mountains.

A departs at 8 o'clock and walks to 10 o'clock, and the distance is 6 2 = 12 kilometers.

The journey is 5 1 = 5 km at 9 o'clock in the evening.

Data for common right-angled triangles.

Connect the AF

af=fcef+fc=ef+af.If and only if, a, f, e three points are collinear, the minimum value = ae, the vertical distance from point A to BC.

ae=3√3/2

It should be ac=bc

Extend BC, AE at H

Be bisects abc, then abe= cbe= hbebe ae, then aeb= heb=90°be=be

abe≌△hbe(asa)

ae=eh=1 2ah, i.e. ah=2ae

acb=∠ach=90°

Then cah=90°- h

cbf=∠hbe=90°-∠h

cah=∠cbf

cah=∠cbf

ac=bcfcb=∠ach=90°

ach≌△bcf(asa)

bf=ah=2ae

Let's make sure that you can write the words separately.

Be bisected by ABC

So 1= 3

Again 1= 2

So 2= 3 (equal substitution).

So bc de (the inner wrong angles are equal, and the two lines are parallel).

Solution: BE bisected ABC

BC Ping DE

Thank you for adopting.