Given a / | a | + B / | B | + C / | C | = 1, calculate the value of | ABC | (BC / | ab | × AC / | BC | × AB / | Ca | Original formula =

Given a / | a | + B / | B | + C / | C | = 1, calculate the value of | ABC | (BC / | ab | × AC / | BC | × AB / | Ca | Original formula =

∵a/|a|+b/|b|+c/|c|=1
ν there are two greater than zero and one less than zero
Qi
|abc|/abc÷(bc/|ab|*ac|bc|*ab/|ca|)
=-1÷1
=-1

Given ABC = 1, find a / AB + A + 1 + B / BC + B + 1 + C / Ca + C+ Given ABC = 1, find the value of a / AB + A + 1 + B / BC + B + 1 + C / Ca + C + 1

If ABC = 1, then
a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)
=a/(ab+a+abc)+b/(bc+b+1)+bc/(cba+bc+b)
=1/(b+1+bc)+b/(bc+b+1)+bc/(1+bc+b)
=(1+b+bc)/(1+b+bc)
=1

Given ABC = 1, find the value of a / AB + A + 1 + B / BC + B + 1 + C / Ca + C + 1 among X = 19-8 times root sign 3 =Under root sign [4 ^ 2 + (root 3) ^ 2-2 * 4 * root 3] =Radical (4-radical 3) ^ 2 =4-radical 3 X-4 = - radical 3 (x-4)^2=3 x^2-8x+13=0 x^2-8x+13=0, So x ^ 2-8x + 15 = 2; x^4-6x^3-2x^2+18x+23 =x^2(x^2-8x+13)+2x^3-15x^2+18x+23 =2x(x^2-8x+13)+x^2-8x+23 =x^2-8x+23 =x^2-8x+13+10=10 So: (x ^ 4-6x ^ 3-2x ^ 2 + 18x + 23) / (x ^ 2-8x + 15) = 10 / 2 = 5 =How to get 2x (x ^ 2-8x + 13) + x ^ 2-8x + 23 is just that. I hope there is no nonsense

If the last formula = 2x (x * 2 - 8x + 13) + 2x + 2 * 2x - 30x + 18x + 23, then 2 * 2x - 30x + 18x is equal to - 8x

Let a, B, C be real numbers, and AB / A + B = 1 / 3, BC / B + C = 1 / 4, CA / C + a = 1 / 5. Find the value of ABC / AB + BC + ca

Because AB / (a + b) = 1 / 3, BC / (B + C) = 1 / 4, CA / (c + a) = 1 / 5, so: (a + b) / AB = 3 (B + C) / BC = 4 (a + C) / AC = 5, that is: 1 / A + 1 / b = 3 1 / B + 1 / C = 4 1 / A + 1 / C = 5, the sum of the three formulas is: 2 (1 / A + 1 / B + 1 / C) = 12, so: 1 / A + 1 / B + 1 / C =

It is known that a, B and C are real numbers, and ab a+b＝1 3，bc b+c＝1 4，ca c+a＝1 5. Find ABC The value of AB + BC + Ca

Take the reciprocal of the known three fractions to get a + B
ab＝3，b+c
bc＝4，c+a
ca＝5，
That is 1
A+1
b＝3，1
B+1
c＝4，1
C+1
a＝5，
Add the three formulas to get 1
A+1
B+1
c＝6，
Results: ab + BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

If a, B and C are real numbers and ABC = 1, then 1 a+ab+1+1 b+bc+1+1 The value of C + Ca + 1 is () A. 1 Two B. 1 Three C. 1 D. 3

∵ a, B, C are all real numbers, and ABC = 1, then AC = 1
b，
The original formula = ABC
a+ab+abc+1
b+bc+1+1
C+1
B+1
=bc
b+bc+1+1
b+bc+1+b
b+bc+1
=bc+b+1
b+bc+1
=1．
Therefore, C

It is known that a, B and C are real numbers, and ab a+b＝1 3，bc b+c＝1 4，ca c+a＝1 5. Find ABC The value of AB + BC + Ca

Take the reciprocal of the known three fractions to get a + B
ab＝3，b+c
bc＝4，c+a
ca＝5，
That is 1
A+1
b＝3，1
B+1
c＝4，1
C+1
a＝5，
Add the three formulas to get 1
A+1
B+1
c＝6，
Results: ab + BC + ca
abc＝6，
That is, ABC
ab+bc+ca=1
6．

0

(a+b+c)^2=1
a^2+b^2+c^2+2(ab+bc+ac)=1
a^2+b^2+c^2>=1/3>=ab+bc+ac

Let the three sides of △ ABC be a, B, C, and prove that ab + BC + Ca ≤ A2 + B2 + C2 < 2 (AB + BC + Ca)

It is proved that: ∵ A2 + B2 ≥ 2Ab, B2 + C2 ≥ 2BC, A2 + C2 ≥ 2Ac, can be obtained by adding 2 (A2 + B2 + C2) ≥ 2Ab + 2BC + 2Ac, A2 + B2 + C2 ≥ AB + BC + ca

What inequalities are commonly used in high school, and there is a process of proof

1. Arithmetic geometric mean inequality
2. Cauchy inequality
3. Ranking inequality
The above is the inequality required by the League examination