ab/bc = 1/2 a/c = 1/2 2a = c → given: ac = 3 → c = 3/a 2a = 3/a 2a² = 3 a² = 3/2 bc/ac = 2/3 b/a = 2/3 b = (2/3) * a b² = (4/9) * a² b² = (4/9) * (3/2) b² = 12/18 b² = 2/3 Recall: 2a = c c = 2a c² = 4a² c² = 4 * (3/2) c² = 6 (a + b + c)² = (a + b + c).(a + b + c) (a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c² (a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc (a + b + c)² = a² + b² + c² + 2.(ab + ac + bc) (a + b + c)² = (3/2) + (2/3) + 6 + 2.(1 + 2 + 3) (a + b + c)² = (13/6) + 18 (a + b + c)² = 121/6 a + b + c = ± √(121/6) a + b + c = ± √(121)/(√6) a + b + c = ± 11/(√6) a + b + c = ± (11√6)/6
It is similar to earlier one. Now you have. ab=1. Then bc=2. And then came=3. Multiplying the first one and the second one you get a^2(bc)=a^2(2)and this is =1×2.@2.So 2a^2=2 So ^2=1. Like that you can get b value and value for c. Also. By adding you get the answer.
bc=2=2ab gives c=2a. 2a^2=3 gives a=sqrt(3/2), which in turn gives b=sqrt(2/3). Finally, c=sqrt(6). Hence, a+b+c=sqrt(3/2)+sqrt(2/3)+sqrt(6)=(2+3+6)/sqrt(6)=11/sqrt(6).
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from Morocco thank you.....at mn 3-04 we had a=1/b we can replace and deduce that abc=a.(1/a)c=sq6....then c=sq6.....then get all values of b=2/sq6 and a=3/sq6 then compute theire summ
the previous handwriting with a marker is much better than the new one.
Thank you 😀
(ab ➖1ab+1) (bc ➖1bc+1) (ca ➖2 ca+1).
ab/bc = 1/2
a/c = 1/2
2a = c → given: ac = 3 → c = 3/a
2a = 3/a
2a² = 3
a² = 3/2
bc/ac = 2/3
b/a = 2/3
b = (2/3) * a
b² = (4/9) * a²
b² = (4/9) * (3/2)
b² = 12/18
b² = 2/3
Recall: 2a = c
c = 2a
c² = 4a²
c² = 4 * (3/2)
c² = 6
(a + b + c)² = (a + b + c).(a + b + c)
(a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c²
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc
(a + b + c)² = a² + b² + c² + 2.(ab + ac + bc)
(a + b + c)² = (3/2) + (2/3) + 6 + 2.(1 + 2 + 3)
(a + b + c)² = (13/6) + 18
(a + b + c)² = 121/6
a + b + c = ± √(121/6)
a + b + c = ± √(121)/(√6)
a + b + c = ± 11/(√6)
a + b + c = ± (11√6)/6
(ab)(bc)(ca)=1×2×3
a²b²c²=6
abc=√6
a=(√6)/bc
a=√6/2
b=(√6)/ac
b=√6/3
c=(√6)/ab
c=√6/1=√6
a+b+c=√6/2+√6/3+√6
=(3√6+2√6+6√6)/6=11√6/6
a+b+c= ± (11/6)√6
It is similar to earlier one. Now you have. ab=1. Then bc=2. And then came=3. Multiplying the first one and the second one you get a^2(bc)=a^2(2)and this is =1×2.@2.So 2a^2=2 So ^2=1. Like that you can get b value and value for c. Also. By adding you get the answer.
Multiplying all equations and taking square root abc = √6
Now
a+b+c = abc (1/bc+1/ca+1/ab)
= √6 (1/2+1/3+1)
= √6×11/6
= 11√6/6
SENSATIONAL!
Shorts..
a/c= 1/2; then a^2= 3/2 or a= √3/√2 then b= √2/√3; c= √6
a+b+c=11/ √ 6
bc=2=2ab gives c=2a. 2a^2=3 gives a=sqrt(3/2), which in turn gives b=sqrt(2/3). Finally, c=sqrt(6).
Hence, a+b+c=sqrt(3/2)+sqrt(2/3)+sqrt(6)=(2+3+6)/sqrt(6)=11/sqrt(6).
from Morocco thank you.....at mn 3-04 we had a=1/b we can replace and deduce that abc=a.(1/a)c=sq6....then c=sq6.....then get all values of b=2/sq6 and a=3/sq6 then compute theire summ
a=√(3/2) b=√(2/3) c=√6
a+b+c= (1/√6)(3+2+6)
a+b+c= 11/√6
Good but you have forgotten the negative value a+b+c=-11sqrt(6)/6
Porque você não considerou a soma negativa? Why didn't you consider the sum negative?