Knowing abc 1 a b c 2 a b c 3, what is the value of 1 ab c 1 1 bc a 1 1 ca b 1?

The original form can be reduced to:

1/(ab+c-1)+1/(bc+a-1)+1/(ca+b-1)

1/(ab+1-a-b)+1/(bc+1-b-c)+1/(ca+1-c-a)

1/[(1-a)(1-b)]+1/[(1-b)(1-c)]+1/[(1-c)(1-a)]

1-c+1-a+1-b)/[(1-a)(1-b)(1-c)]

1/[(1-a)(1-b)(1-c)]

The denominator is calculated on the same floor, and after multiplication, it is [abc+(ab+bc+ca)-(a+b+c)+1].

where ab+bc+ca=[(a+b+c) 2-(a 2+b 2+c 2)] 2=1 2, and the denominator is 1 2

So the final result is 2

Knowing abc=1, a+b+c=2, a 2+b 2+c 2=3, then (a+b+c) 2=2 2 a 2+b 2+c 2+2ab+2ac+2bc=4

ab+bc+ac=[4-(a^2+b^2+c^2)]/2=(4-3)/2=1/2 (ab+bc+ac)^2=1/4 aabb+bbcc+aacc=1/4-2*(a+b+c)=1/4-2*2=-15/4

Through the differentiation simplification, and abc = 1, a + b + c = 2, a 2 + b 2 + c 2 = 3, ab + bc + ac = 1 2, aabb + bbcc + aacc = -15 4 substitution, molecule 1) = (ab + c-1) * (bc + a - 1) = abbc + aab-ab + bcc + ac-c-bc-a + 1 = b + aab-ab + bcc + ac - c - bc-a + 1

Molecule 2) = (BC+A-1)*(Ca+B-1)=C+BBC-BC+AAC+AB-A-AC-B+1

Molecule 3) = (ab+c-1))*ca+b-1)=a+abb-ab+acc+bc-c-ac-b+1

Molecule = (Molecule 1) + (Molecule 2) + (Molecule 3).

3-(a+b+c)-(ab+bc+ac)+aab+abb+bbc+bcc+aac+acc

3-2-1/2+ab*(a+b)+bc*(b+c)+ac*(a+c)

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

1/2+2*(ab+bc+ac)-3abc

Denominator = (ca+b-1)*(numerator 1).

ca+b-1)*(b+aab-ab+bcc+ac-c-bc-a+1)

aa+bb+cc)+(aabb+bbcc+aacc)-(acc+aac)-(abb+aab)-(bcc+bbc)+1

3-15/4+1-ac*(a+c)-ab*(a+b)-bc(b+c)

1/4-ac*(2-b)-ab*(2-c)-bc*(2-a)

1/4-2*(ac+bc+ab)+3abc

Because aabb+bbcc+aacc=-15 4 is in the range of complex numbers, the result: 1 (ab+c-1)+1 (bc+a-1)+1 (ca+b-1) =(-3 2) (9 4) =-2 3

The substitution value promotion plan is ridiculous and calculating. Namely.

Bitwise operations are calculated first &&

So grinding laughter -1||1

That is, 1

1 a+1 b=31 a=3-1 b=1 b=1 1 c+1 b=5 1 c+3-1 b=51 c-1 b=2 11 b+1 c=4 21+2 2 c=61 c=3 c=1 32-1 2 b=21 b=1 b=1 b=1 1 b=1 1 a=1 b=3 1 a=3-1=2a=1 2abc=1 2*1*1 3=1 6ab+bc+ca=1 2*1+1*1 3+1 2*1 3=1 so abc ab+bc+c....

1 a+1 b=1,1 disguised brother b+1 c=2,1 nuclear attack a+1 c=3. 2/a+2/b+2/c=6

1/a+1/b+1/c=3

abc (ab+ac+bc) (divide the denominator by abc)1 (ab abc+ac abc+bc abc)1 (1 c+1 b+1 a).

The answer is -2 3

Solution: Knowing abc=1,a+b+c=2,a2+b2+c=3,then.

a+b+c）^2=2^2

a^2+b^2+c^2+2ab+2ac+2bc=4

ab+bc+ac=[4-(a^2+b^2+c^2)]/2=(4-3)/2=1/2

ab+bc+ac)^2=1/4

aabb+bbcc+aacc=1/4-2*(a+b+c)=1/4-2*2=-15/4

Scores. Simplify, and substitute abc=1,a+b+c=2,a 2+b 2+c 2=3,ab+bc+ac=1 2,aabb+bbcc+aacc=-15 4, molecule 1)=(ab+c-1)*(bc+a-1)=abbc+aab-ab+bcc+ac-c-bc-a+1

b+aab-ab+bcc+ac-c-bc-a+1

Molecule 2) = (BC+A-1)*(Ca+B-1)=C+BBC-BC+AAC+AB-A-AC-B+1

Lead molecule 3) = (ab + c-1)) * ca+b-1) = a + abb-ab + acc + bc-c-ac-b + 1

Molecule = (Molecule 1) Burn well + (Molecule 2) + (Molecule 3).

3-(a+b+c)-(ab+bc+ac)+aab+abb+bbc+bcc+aac+acc

3-2-1/2+ab*(a+b)+bc*(b+c)+ac*(a+c)

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

1/2+2*(ab+bc+ac)-3abc

Denominator. (ca+b-1)*(numerator 1)=(ca+b-1)*(b+aab-ab+bcc+ac-c-bc-a+1)=1+aa-a+cc+aacc-acc-c-aac+ac+bb+aabb-abb+bbcc+1-bc-bbc-ab+b-b-aab+ab-bcc-ac+c+bc+a-1

aa+bb+cc)+(aabb+bbcc+aacc)-(acc+aac)-(abb+aab)-(bcc+bbc)+1

3-15/4+1-ac*(a+c)-ab*(a+b)-bc(b+c)

1/4-ac*(2-b)-ab*(2-c)-bc*(2-a)

1/4-2*(ac+bc+ab)+3abc

Because aabb+bbcc+aacc=-15 4, which is in the range of complex numbers, the result:

1/(ab+c-1)+1/(bc+a-1)+1/(ca+b-1)

3 2) Defeat (9 4).

wxywxq：

Hello, the answer is -2 3

Solution: Knowing abc=1,a+b+c=2,a2+b2+c=3,then.

a+b+c）^2=2^2

a^2+b^2+c^2+2ab+2ac+2bc=4

ab+bc+ac=[4-(a^2+b^2+c^2)]/2=(4-3)/2=1/2

ab+bc+ac)^2=1/4

aabb+bbcc+aacc=1/4-2*(a+b+c)=1/4-2*2=-15/4

Through the differentiation simplification, and abc = 1, a + b + c = 2, a 2 + b 2 + c 2 = 3, ab + bc + ac = 1 2, aabb + bbcc + aacc = -15 4 substitution, (molecule 1) = (ab + c - 1) * (bc + a - 1) = abbc + aab-ab + bcc + ac-c - bc-a + 1

b+aab-ab+bcc+ac-c-bc-a+1

Molecule 2) = (BC+A-1)*(Ca+B-1)=C+BBC-BC+AAC+AB-A-AC-B+1

Molecule 3) = (ab+c-1))*ca+b-1)=a+abb-ab+acc+bc-c-ac-b+1

Molecule = (Molecule 1) + (Molecule 2) + (Molecule 3).

3-(a+b+c)-(ab+bc+ac)+aab+abb+bbc+bcc+aac+acc

3-2-1/2+ab*(a+b)+bc*(b+c)+ac*(a+c)

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

1/2+2*(ab+bc+ac)-3abc

Denominator = (ca+b-1)*(numerator 1)=(ca+b-1)*(b+aab-ab+bcc+ac-c-bc-a+1)=1+aa-a+cc+aacc-acc-c-aac+ac+bb+aabb-abb+bbcc+1-bc-bbc-ab+b-b-aab+ab-bcc-ac+c+bc+a-1

aa+bb+cc)+(aabb+bbcc+aacc)-(acc+aac)-(abb+aab)-(bcc+bbc)+1

3-15/4+1-ac*(a+c)-ab*(a+b)-bc(b+c)

1/4-ac*(2-b)-ab*(2-c)-bc*(2-a)

1/4-2*(ac+bc+ab)+3abc

Because aabb+bbcc+aacc=-15 4, which is in the range of complex numbers, the result:

1/(ab+c-1)+1/(bc+a-1)+1/(ca+b-1)

Plump King.

abc = 1 so b = 1 ac

ab=1/c

bc=1/a

So. Original formula = a (1 c+a+1) + (1 ac) (1 a+1 ac+1) + c (ac+c+1).

The first equation multiplies c up and down

The second equation is multiplied by AC up and down

So =ac (ac+c+1)+1 (ac+c+1)+c (ac+c+1).

ac+c+1)/(ac+c+1)

I noticed that a+b+c is greater than or equal to 3

This question is wrong, I asked others, even if it was a reference.