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It is known that the three sides of an unequal triangle ABC are integers a, B and C respectively, and the length of C can be obtained if the square of a + the square of b-4a-6b + 13 = 0
According to the question condition a square + b square - 4A - 6B + 13 = 0, we can get (AA - 4A + 4) + (BB - 6B + 9) = 0, that is, (A-2) square + (B-3) square = 0, so a = 2, B = 3A, B C is the three sides of the triangle, so B - a < C < B + A, i.e. 1 < C < 5
If the positive integers a, B and C satisfy the inequality (a square + b square + C square + 48) less than 6A + 4A + 12C, then (1 / A + 1 / B + 1 / C) has the ABC power=
Positive integers a, B and C satisfy the following conditions:
a²+b²+c²+48<6a+4b+12c
(a²-6a+9)+(b²-4b+4)+(c²-12c+36)-1<0
(a-3)²+(b-2)²+(c-6)²<1
a. B and C are positive integers, so (A-3) 2 + (b-2) 2 + (C-6) 2 must be integers greater than or equal to 0
So, (A-3) 2 + (b-2) 2 + (C-6) 2 = 0
a-3=0,b-2=0,c-6=0
a=3,b=2,c=6
（1/a+1/b+1/c)^abc
=(1/3+1/2+1/6)^36
=1^36
=1
Given a ^ 2 + B ^ 2 + C ^ 2 + 49 = 4A + 6B + 12C, find the value of (1 / A + 1 / B + 1 / C) ^ ABC As the title
2 + B ^ 2 + C ^ 2 + 49-4a-6b-12c = 0 (a ^ 2-4a + 4) + (b ^ 2-6b + 9) + (b ^ 2-6b + 9) + (C ^ 2-12c + 36) = 0 (A-2) ^ 2 + (B-3) ^ 2 + (C-6) ^ 2 = 0, the sum is equal to 0. If one is greater than 0, then the other is less than 0, which is not true. Therefore, two are equal to 0A = 2B = 3C = 6 (1 / A + 1 / B + 1 / C) ^ ABC = (1 / 2 + 1 / 3 + 1 / 6 / 6 / 6 / 6) ABC = (1 / 2 + 1 / 3 + 1 / 6 / 6 / 6 / 6 / 6 / 6, this is not true. So this is not true. So and
Given that a, B, C are integers, and a + B + C + 48 is less than 4A + 6B + 12C, find the value of (1 / A + 1 / B + 1 / C) to the ABC power
a²+b²+c²+48
=(a-2)²+(b-3)²+(c-6)²-1+ 4a+6b+12c< 4a+6b+12c
=(a-2)²+(b-3)²+(c-6)²
If a, B, C are the three sides of △ ABC, and A2 + B2 + C2 = AB + AC + BC, then △ ABC is () A. Isosceles triangle B. Right triangle C. Equilateral triangle D. Isosceles right triangle
0
0
（1）∵a2+b2+c2+200=12a+16b+20c，
∴（a2-12a+36）+（b2-16b+64）+（c2-20c+100）=0，
That is (a-6) 2 + (B-8) 2 + (C-10) 2 = 0
A-6 = 0, B-8 = 0, C-10 = 0, that is, a = 6, B = 8, C = 10, and 62 + 82 = 100 = 102,
∴a2+b2=c2，
The △ ABC is a right triangle
（2）（a3-a2b）+（ab2-b3）-（ac2-bc2）=0，a2（a-b）+b2（a-b）-c2（a-b）=0，
∴（a-b）（a2+b2-c2）=0
A-B = 0 or A2 + b2-c2 = 0 or (a-b) (A2 + b2-c2) = 0,
The triangle ABC is an isosceles triangle, a right triangle or an isosceles right triangle
It is known that a, B and C are the three side lengths of △ ABC, and satisfy the requirements of a 2 ＋ B C ab-ac-bc = 0, try to judge the shape of △ ABC
2a²＋2b²＋2c²-2ab-2ac-2bc＝0
(a-b)²+(b-c)²+(c-a)²=0
a-b=0、b-c=0、c-a=0
a=b=c
So Δ ABC is an equilateral triangle
It is known that a, B, C are the three side lengths of △ ABC, and a? + B? + C? - AB BC AC = 0, try to judge the shape of △ ABC
Answer: the shape of triangle ABC is an equilateral triangle
Multiply the left and right sides of the original formula by 2
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0
That is (a-b) ^ 2 + (B-C) ^ 2 + (A-C) ^ 2 = 0
So A-B = 0, B-C = 0, a-c = 0
So a = b = C
So the shape of triangle ABC is an equilateral triangle
Note: you can ask if you have any questions
It is known that a, B and C are the three side lengths of △ ABC, and a 2 + B C AB BC AC = 0 is used to judge the shape of △ ABC
According to the properties of the equation, it is concluded that 2A 2 + 2B 2 + 2C 2 - 2ab-2bc-2ac = 0
The deformation is as follows: (a 2 - 2Ab + B 2) + (B 2 - 2BC + C 2) + (a 2 - 2 AC + C 2) = 0
That is: (a-b) 2 + (B-C) 2 + (A-C) 2 = 0
So: A-B = 0
b-c=0
a-c=0
The solution is: a = b b b = C a = C
So a = b = C
So △ ABC is an equilateral triangle
It is known that a, B and C are the three side lengths of △ ABC, and satisfy the requirements of a B = AC BC. Try to judge the shape of the eating triangle
∵a2-b2=ac-bc
∴（a-b）（a+b）=c（a-b）
∴（a-b）（a+b-c）=0
∵ a, B, C are the three sides of ∵ ABC,
∴a+b-c≠0
∴a-b=0
∴a=b
The △ ABC is an isosceles triangle
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