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Given that a = 2013 parts of 2012, B = 2012 parts of 2012, C = 1, find the algebraic formula 3 (AB + BC) minus 3 (AB minus AC) minus 4ac minus 3bC
3(ab+bc)-3(ab-ac)-4ac-3bc=3ab+3bc-3ab+3ac-4ac-3bc=-ac=-2012/2013
Given a + x 2 = 2007, B + x 2 = 2008, C + x 2 = 2009, ABC = 1, try to find (a in BC) + (B in CA) - a ′ 1-B ′ 1-C ′ 1
(a in BC) + (B in CA) - a ′ 1-B ′ 1-C ′ 1
Can the above symbol be expressed in Chinese characters
Given a + x2 = 2003, B + x2 = 2004, C + x2 = 2005, and ABC = 6012, find a bc+b ca+c ab−1 a−1 b−1 The value of C
B-A = 1, C-B = 1, C-A = 2, original formula = A2 + B2 + B2 + c2bc - (1a + 1b + 1c) = A2 + B2 + c2bc - (1a + 1b + 1c) = A2 + B2 + c2abc + AC + ababc = A2 + B2 + c2abc + AC + ababc = A2 + B2 + C2 + c2bc + AC + ababc = A2 + B2 + C2 + c2bc + AC + ababc = A2 + B2 + C2 − BC − ababc = a (a − C) + B (B − a) + C (C − b) ABC,
If ABC = 1, solve the equation about X: 1 in X + A + AB + 1 in X + B + BC + 1 in X + C + Ca = 2005
The equation can be changed into (change 1 in denominator of 1,2 into ABC)
x/abc+a+ab+x/abc+b+bc+x/1+c+ca=2005
It is equivalent to (in which the denominator distributions of 1 and 2 terms propose a and B respectively)
x/a（bc+1+b）+x/b(ac+1+c)+x/1+c+ca=2005
It is equivalent to (change the 1 in the 1 term of the formula into ABC)
x/a（bc+abc+b）+x/b(ac+1+c)+x/1+c+ca=2005
It is equivalent to (put B in brackets in Item 1)
x/ab（c+ac+1）+x/b(ac+1+c)+x/1+c+ca=2005
Equivalent to (General Division)
x＋ax＋abx/ab（1+c+ca）＝2005
Equivalent to (propose x)
x（1＋a＋ab）/ab（1+c+ca）＝2005
Equivalent to (replace 1 in molecule with ABC)
x（abc＋a＋ab）/ab（1+c+ca）＝2005
Equivalent to (molecule proposed a)
xa（bc＋1＋b）/ab（1+c+ca）＝2005
It is equivalent to (replace 1 in the molecule with ABC)
xa（bc＋abc＋b）/ab（1+c+ca）＝2005
Equivalent to (molecule proposed b)
xab（c＋ac＋1）/ab（1+c+ca）＝2005
Equivalent to (approximately)
x＝2005
Note, equivalent to the following brackets is the next step of the description, where 1 item refers to X / 1 + A + ab
The two terms refer to X / 1 + B + BC
The three terms refer to X / 1 + C + ca
Then we know that the equation ABC = 1 1+a+ab+x 1+b+bc+x The solution of 1 + C + AC = 2012 is______ .
In this paper, we analyze the equation of X (x1 + A + AB + 11 + AB + 11 + B + B + B + 11 + C + 11 + C + AC) x = 2012, namely (C1 + C + AC + 11 + C + 11 + B + 11 + B + 11 + B + BC = 11 + B + 1A = a1 + A + A + AB, (11 + B + BC = 11 + B + 1A = a1 + A + A + AB, 11 + B + BC = AC 1 + C + AC, 9 the equation about X is X1 + A + AB + X1 + B + BC + X1 + C + AC = 2012, namely (11 + A + AB + 11 + B + 11 + 11 + C + AC) x = 2012, namely (C1 + C + AC + AC + AC + AC + 11 + 11 + 11 + 11 + 11 + 11 + 11 +
Given a + x 2 = 2009, B + x 2 = 2010, C + x 2 = 2011, and ABC = 6030, calculate the value of a / BC + B / Ca + C / AB-1 / A-A / B-A / C
a+1=b；b+1=c；a+2=ca/bc+b/ca+c/ab-1/a-1/b-1/c=(a^2+b^2+c^2-bc-ac-ab)/abc=(a^2+b^2+c^2-bc-ac-ab)/6030=(a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac)/(6030*2)=((a-b)^2+(a-c)^2+(b-c)^2)/(6030*2)=(1+4+1)/(6030*2)=...
Known a + x ^ 2 = 2, B + x ^ 2 = 2010, C + x ^ 2 = 2011, and ABC = 1. Find the value of the algebraic expression a / BC + B / AC + C / AB-1 / A-1 / B-1 / C
A / BC + B / AC + C / AB = (a? + B? + C?) / ABC1 / A + 1 / B + 1 / C = (AB + BC + AC) / ABCA + X? = 2? Should = 2009? The original formula = (a? + B? + C?) / ABC - (AB + BC + AC) / ABC = (a? + B? + C? - BC AC AB) / ABC = ((a-b)? + (a -...)
If a = 2009 in 2010, B = 2010 in 2011, C = 2011 in 2011, then the size relationship of ABC is
C>B>A
ABC = 1, solve the equation of X: (x / 1 + A + AB) + (x / 1 + B + BC) + (x / 1 + C + Ca) = 2009
A + B + A + B = a + B + B = a + B + B = a + B + B = a + B + A + B + B = a + B + B = a + B + B = a + B + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + A + B + B +
Given ABC = 1, solve the equation about X, (2aX / AB + A + 1) + (2bx / BC + B + 1) + (2cx / Ca + C + 1) = 1
2x[(a/ab+a+1)+(b/bc+b+1)+(c/ca+c+1)]=1
2x[a/（ab+a+abc）+b/（bc+b+1）+c/（ca+c+1）]=1
2x [ 1/（b+1+bc）+b/（bc+b+1）+c/（ac+c+1）]=1
2x （1+b）/（bc+b+1）+c/（ac+c+1） =1
2x（abc+b）/（bc+b+abc）+c/（ac+c+1） =1
2（ac+1）/（c+1+ac）+c/（ac+c+1）=1
2x（ac+c+1）/（ac+c+1）=1
2x×1 =1
x=1/2
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