Knowing that a b c satisfies a 2b 7, b 2c 1, c 6c 17, find the value of a b c

c²-6c=17

c²-6c+9=26

c-3)²=26

c = 3 + 26 or 3 - 26

b = 1 + 2 c = 7 + 2 26 or 7-2 26

a = 7-2b = -2 26 or 2 26

Because a is greater than or equal to 0, a = -2 26 is rounded to c=3 + 26 and b = 7 + 2 26

So a = (104) (1 4), b = 7-2 26, c = 3- 26a + b + c = (104) (1 4) + 10 - 3 26

First find c to get two values, bring in to find b is also two values, according to b to find a, because a2 = 7-2b must be non-negative, reverse b and c only one value to meet the requirements, get two values of a, and then find the result of abc addition, there are two values.

I made a mistake. a 2+2b=7,b 2-2c=-1,c 2-6a=-17 like that.

a^2+2b)+(b^2-2c)+(c^2-6a)=7-1-17=-11

a^2-6a+9)+(b^2+2b+1)+(c^2-2c+1)=0a-3)^2+(b+1)^2+(c-1)^2=0a-3=0b+1=0

c-1=0a=3

b=-1c=1

a+b+c=3

a +b Qi scattered branch = (a + b) 2-2ab, and a + b = 6-c, ab = 2 c

So a +b high sensitivity = (6-c) 2-4 c and a +b +c =25, so (6-c) 2-4 c + c = 25, solve the equation to get c =

Separate the source of the excavation into the solution a, b

This is the matching method used, and the constant term is put together:

a²-6a+b²+2b+c²-2c=-11,a²-6a+9+b²+2b+1+c²-2c+1=0(a²-6a+9)+(b²+2b+1)+(c²-2c+1)=0(a-3)²+b+1)²+c-1)²=0,

a -6a+b +2b+c -2c=-11, i.e.: a -6a+b +2b+c -2c+11=0a -6a+9+b +2b+1+c -2c+1=0(a-3) +b+1) +c-1) =0 So, a-3=0, b+1=0, c-1=0

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

Answer: (1) To the left of the equal sign.

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

log(2)((ab+b^2+bc+a^2-ac+ab+ac-bc-c^2)/ab)

log(2)((ab+bc-ac+ab+ac-bc)/ab)=log(2)((2ab)/ab)

log(2)2

1=Right. So the original formula is true.

2) i.e. (b+c) a=3, a+b-c=4, c=3a-b, c=a+b-4

So 3a-b=a+b-4, i.e., a=b-2, is substituted in c=2b-6 and gets from a2+b2=c2.

b-2)^2+b^2=(2b-6)^2

i.e. b 2-10b+16 = 0, (b-8) (b-2) = 0, so b = 8 or b = 2

When b = 8, a = 6, c = 10

When b=2, a=0 and the denominator is 0. Abandon it.

So a=6, b=8, c=10

is composed of a(1 b+1 c) + b(1 a+1 c) + c(1 a+1 b) = -3

It can be obtained: (a+b+c)*(1 a+1 b+1 c)=0 a, b,c cannot be zero, abc≠0

Therefore, it gets: (a+b+c)*(bc+ca+ab)=0 a+b+c=0, or bc+ca+ab=0

again a 2 + b 2 + c 2 = 1

If bc+ca+ab=0, then.

a+b+c)^2=1-2(bc+ca+ab)=1.

a+b+c=±1.

a+b+c=0,1,-1.

∵ a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3

a(1/b+1/c)+a/a+b(1/a+1/c)+b/b+c(1/a+1/b)+c/c=-3+3=0

a+b+c)(1/a+1/b+1/c)=0∵ a≠0,b≠0,c≠0

abc≠0a+b+c)(ab+bc+ac)=0, i.e. (a+b+c)=0 or (ab+bc+ac)=0, when (ab+bc+ac)=0.

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

So a+b+c= 1

In summary, there are three values of a+b+c: -1, 0, 1

Analysis: Equation a +b +c -ab-6b-6c + 21 = 0 multiplied by 4 on both sides to obtain:

4a +4b +4c -4ab-24b-24c + 84=0 formula burning knowledge available:

4a -4ab + b +3 (b -8b + 16) + 4 (c nuclear phase -6c + 9) = 0

i.e. (2a-b) +3(b-4) +4(c-3) =0 For the upper limb modification to be established, it must be made so that:

2a-b=0，b-4=0，c-3=0

Solution: b=4, c=3, a=2

a +b +c -ab-6b-6c+21=0a-b Zheng Jian2) +3 4(b-4) +c-3) =0a-b 2) +3 4(b-4) +c-3) Yun Chi = 0 so only (a-b 2) =0 3 4(b-4) =0 (c-3) shouting pants = 0

b=4 c=3 a=2

So a=2 b=4 c=3