It is known that the three sides of △ ABC are a, B and C, and a + B = 4, ab = 1, C= 14. Try to determine the shape of △ ABC

∵a+b=4，ab=1，
∴a2+b2=（a+b）2-2ab=16-2×1=14，
∵c=
14，
∴c2=14，
∴a2+b2=c2，
The △ ABC is a right triangle

It is known that the three sides of a triangle are ABC, where both sides of AB satisfy the root sign (a? - 12a + 36) + root sign (B-8) = 0 Find this triangle (1) The value range of the third side C (2) The value range of maximum edge C (3) The range of minimum edge C

(a ^ 2-12a + 36) + radical (B-8) = 0
If (a-6) ^ 2 + radical (B-8) = 0, the above equation will hold,
Only a-6 = 0 and B-8 = 0
A = 6 and B = 8
(1) The value range of the third side C
2 (2) the value range of maximum edge C
8 (3) the range of minimum edge C
Two
Question:
Sorry, A-B = 0, B-8 = 0, a = 6, B = 8, I don't quite understand these two, can you elaborate?
Answer:
(a ^ 2-12a + 36) + root sign (B-8) = 0 (a-6) ^ 2 > = 0 root (B-8) > = 0 and the two addition = 0, so both = 0 (a-6) ^ 2 = 0, a-6 = 0, a = 6, radical (B-8) = 0, B-8 = 0, B = 8
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Given that the three sides of the triangle ABC are a, B, C, and satisfy the requirements of 4A? - 12a + 9 + radical B-2 + | 2c-5 | = 0, then a = b = C = the shape of the triangle ABC is__ triangle

a=1.5
B=2
c=2.5
right triangle

It is known that the three sides of the triangle are ABC, where both sides of AB satisfy the root sign (a-12a + 36) + the root sign (B-8) = 0, and find the value of the longest side C

(a-12a + 36) + radical (B-8) = 0 (a-6) + radical (B-8) = 0 to make the above equation hold, only if a-6 = 0 and B-8 = 0, a = 6 and B = 8, the longest side C must be greater than a, B C > 8, both sides of the triangle and greater than the third side, then a + b > C C C

In the triangle ABC, if we know that a? = B? + C? + BC, 2b = 3C, a = 3 root sign 19, then the area of triangle ABC?

From the cosine theorem, we get a 2 = B 2 + C 2 - 2 bccosa. By substituting a 2 = B 2 + C 2 + BC into the above formula, we can get B 2 + C 2 + B C = B 2 + C 2 - 2 bccosa. We can get cosa = - BC / 2BC = - 1 / 2 because a is the inner angle of the triangle, so a = 120 ° from 2B = 3C, B = 3C / 2

In ABC, a, C is a triangle (a-b+c)2-2|c-a-b|．

∵ in ᙽ ABC, a, B, C are the three sides of a triangle,
∴a-b+c＞0，c-a-b＜0，
The original formula = A-B + C-2 [- (C-A-B)]
=a-b+c+2c-2a-2b
=-a-3b+3c．

In △ ABC, a, B and C are the three sides of a triangle. Under the root sign, (a-b-c) 2-2 | C-A-B | 3 | B-C + a are simplified|

In △ ABC, a, B, C are the three sides of a triangle, and the sum of any two sides is greater than the third side
Under the radical sign (a-b-c) - 2|c-a-b| + 3|b-c + A|
=b+c-a-2(a+b-c)+3(b+c-a)
=2b+6c-6a

Given the position of real number ABC on the number axis as shown in the figure, simplify the root sign a-a-c | and the root sign (C-B) | - | - B|

A < 0, so | a | = - A
A > C means a-c > 0, so | a-c | = a-c
C < B means C-B < 0, so | C-B | = - (C-B) = b-c
B > 0 is - B < 0, so | - B | = - (- b) = B
The original formula = |a | - | a-c | + | C-B | = - A - (A-C) + (B-C) - B = - 2A

Given that the corresponding points of a, B and C on the number axis are shown in the figure, the result of simplifying | a + B | + C-B - radical (A-C) 2 is ABC position as follows: c a 0 b

|A + B | + | C-B | - radical (A-C)
=a+b+b-c-(a-c)
=a+b+b-c-a+c
=2b

If the position of ABC on the number axis is as shown in the graph, then the root sign of the algebraic formula a 2 - | a + B | + root sign (- a) 2 + | B + C | B ﹤ a ﹤ 0 ﹤ C

According to the meaning of the title
A<0
a+b<0
-a>0
B + C < 0 (consider the case that the absolute value of B is greater than that of C)
So the original formula = - A + A + B + (- a) - (B + C)
=b-a-b-c
=-a-c

It is known that the three sides of △ ABC are a, B and C, and a + B = 4, ab = 1, C= 14. Try to determine the shape of △ ABC