Known: a + x ^ 2 = 2000, B + x ^ 2 = 2001, C + x ^ 2 = 2002, and ABC = 2, find the value of a / BC + C / AB + B / ac-1 / A-1 / B-1 / C

Known: a + x ^ 2 = 2000, B + x ^ 2 = 2001, C + x ^ 2 = 2002, and ABC = 2, find the value of a / BC + C / AB + B / ac-1 / A-1 / B-1 / C

=(a²＋b²＋c²)/abc－(ab＋bc＋ca)/abc=[a²＋b²＋c²－ab－bc－ca]/2=(1/2)[(a²－ab)＋(b²－bc)＋(c²－ca)]=(1/2)[a(a－b)＋b(b－c)＋c(c－a)]=(1/2)[－a－b＋2c]=(1/2)[(c－a)＋(c－b)]=3/2.

It is known that a + x 2 = 1999, B + x 2 = 2000, C + x 2 = 2001, and ABC = 24, find a / BC + B / Ca + C / AB-1 / A-1 / B-1 / C There has to be a process

a+1=b b+1=ca+2=c a/bc+b/ac+c/ab-1/a-1/b-1/c=(a²+b²+c²-bc-ac-ab)/abc=[(a²-ac)+(b²-ab)+(c²-bc)]/abc=[a(a-c)+b(b-a)+c(c-b)]/abc=(-2a+b+c)/abc=[(b-a)+(c-a)]/abc=(1+2)/abc=3/a...

It is known that: the square of a + x = 2007, the square of B + x = 2008, the square of C + x = 2009, and ABC = 6027 Let's see clearly: and ABC = 6027 is not the known: the square of a + x = 2008, the square of B + x = 2007, the square of C + x = 2006, and ABC = 24, find C in BC + AB in C + AC in B-A in 1-B in 1-c Please don't give me the wrong answer I'm saying once it's not ABC = 24, please don't give me that answer

According to the title, a, B, C are three connected natural numbers. Because we can't find any three triple natural numbers, the product is equal to 6027, so there is no solution to this problem

It is known that a + x = 2008, B + x = 2009, C + x = 2010, and ABC = 24 Try to find the value of fraction AB C + AC B + BC a-a-b-c

a+x²-(b+x)²=a-b=-1
In the same way
b-c=-1
c-a=2
The original formula = (C? + B? + a? - AB BC CA) / ABC
=(2a²+2b²+2c²-2ab-2bc-2ac)/2abc
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2abc
=[(a-b)²+(b-c)²+(c-a)²]/2abc
=(1+1+4)/(2×24)
=1/8

Given a + x = 2007, B + x = 2008, C + x = 2009, ABC = 1, try to find (a in BC) + (B in CA) - a ′ 1-B ′ 1-C ′ 1

What is the meaning of the symbol above (a in BC) + (B in CA) - a ′ 1-B ′ 1-C ′ 1, which can be expressed in Chinese characters
Please accept if you are satisfied

A + x ^ 2 = 2007, B + x ^ 2 = 2008, C + x ^ 2 = 2009, ABC = 6027

a-b=-1
b-c=-1
a-c=-2
a/bc+b/ac+c/ba-1/a-1/b-1/c
All points
=(a^2+b^2+c^2-ab-bc-ca)/abc
Molecular = (2a ^ 2 + 2B ^ 2 + 2C ^ 2-2ab-2bc-2ac) / 2
[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)]/2
=[(a-b)^2+(b-c)^2+(c-a)^2]/2
=(1+1+4)/2=3
Denominator = ABC = 6027
So the original formula = 3 / 6027 = 1 / 2009

If a = x + 2013, B = x + 2012, C = x + 2011, what is the value of the algebraic formula a squared + B squared + C squared AB BC Ca Hurry!

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

We know that a = 1 / 2012x + 2012, B = 1 / 2011, C = 1 / 2012 + 2013

The original formula = a 2 + B 2 + C 2 - AB BC CA
=(1/2)[2a²＋2b²＋2c²－2ab－2bc－2ca]
=(1/2)[(a²－2ab＋b²)＋(b²－2bc＋c²)＋(c²－2ac＋a²)]
=(1/2)[(a－b)²＋(b－c)²＋(c－a)²]
Because A-B = - 1, B-C = - 1, C-A = 2, then:
The original formula = (1 / 2) [1 + 1 + 4] = 3

Given that a = 2012x + 2011, B = 2012x + 2012, C = 2012x + 2013, calculate the square of a + the square of B + the square of C - AB BC ca Value of, use now! 100

The original formula = a 2 + B 2 + C 2 - AB BC CA
=(1/2)[(a－b)²＋(b－c)²＋(c－a)²]
Because A-B = - 1, B-C = - 1, C-A = 2, then:
The original formula = (1 / 2) [1 + 1 + 4] = 3

Given that a = 1 / 5 x + 2011, B = 1 / 5 x + 2012, C = 1 / 5 x + 2013, find the value of the algebraic formula a squared + B squared + C squared AB AC BC

Solution;
a²+b²+c²-ab-ac-bc
=a²-ab+b²-bc+c²-ac
=a(a-b)+b(b-c)+c(c-a)
=a*(-1)+b*(-1)+2c
=-a-b+2c
=c-a+c-b
=2+1
=3