Given A-B = 3 and B-C = 2, we can find the value of a 2 + B 2 + C 2 - AB BC ca

Given A-B = 3 and B-C = 2, we can find the value of a 2 + B 2 + C 2 - AB BC ca

∵a﹣b=3,b﹣c=2,
A-B + B-C = 5, that is, a-c = 5,
∴（a﹣b）2=9,（b﹣c）2=4,（a﹣c）2=25,
That is, a2-2ab + B2 = 4, ①
b2﹣2bc+c2=9,②
a2﹣2ac+c2=25．③
① + 2 + 3,
a2﹣2ab+b2+b2﹣2bc+c2+a2﹣2ac+c2=4+9+25,
That is, 2 (A2 + B2 + c2-ab-bc-ac) = 38,
∴a2+b2+c2﹣ab﹣bc﹣ac
=19．
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Given A-B = 3, B-C = 5, find the value of a 2 + B + c-ab-bc-ca

∵ A-B = 3, B-C = 5
∴a-c=8
a²+b²+c²-ab-bc-ca
=1/2(a²-2ab+b²)+1/2(b²-2bc+c²)+1/2(c²-2ca+a²)
=1/2(a-b)²+1/2(b-c)²+1/2(c-a)²
=1/2×3²+1/2×5²+1/2×8²
=1/2×9+1/2×25+1/2×64
=1/2×(9+25+64)
=1/2×98
=49
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Given that A-B = 2, B-C = 3, find the value of a? + B? + C? - AB BC ca Given that the polynomial (MX + 4) (2-3x) does not contain x term after expansion, find the value of M Please! Solve!

(1) Because A-B = 2, B-C = 3, a-c = (a-b) + (B-C) = 5A? + B? + C? - AB BC CA = (1 / 2) [(a-b) 2 + (B-C) 2 + (A-C) 2] = (1 / 2) (2? + 3? + 5?) = 19 (2) (MX + 4) (2-3x) = 2mx + 8-3mx? - 12x = - 3mx? +

Given that A-B = B-C = 3 / 5, a 2 + B 2 + C 2 = 1, what is the value of AB + BC + Ca

A-B three times + B-C three times + C-A three times
a²+b²+c²=1
that will do

Let a, B, C be positive numbers, and prove: BC a+ca b+ab c≥a+b+c．

Proof: ∵ 2 (BC)
a+ac
b+ab
c）
=（bc
a+ac
b）+（bc
a+ab
c）+（ac
b+ab
c）
≥2
abc2
ab+2
acb2
ac+2
bca2
BC
=2c+2b+2a，
∴bc
a+ac
b+ab
c≥a+b+c
If and only if a = b = C, the equal sign holds

Example: A, B, C belong to R. it is proved that a + B + C is greater than or equal to ab + BC + ca

Usually, the two sides of an inequality are multiplied by two at the same time
2. The familiar inequality is obtained by adjusting the two sides of the inequality
(this method 1 or adjust to the side of the unequal sign 2 or try to make one side become what you want, and the other side will talk about it last.)
3. Several times of mean inequality
4. Sometimes it can be changed into fraction
5. Sometimes you can write more on "1"

In the triangle ABC, the angle c is equal to the vertical square of 90 ° ab In the RT triangle ABC, the angle c is equal to 90 ° and the vertical square of AB intersects BC at D and ab crosses AB at E Angle CAD: angle DAB = 1:2, calculate the degree of angle B Hurry!

∵∠CAD:∠DAB=1:2
ν if ∠ CAD = x, then ∠ DAB = 2x
And divide AB vertically
∴DA=DB
∴∠B=∠DAB=2x
And ∠ CAD + ∠ DAB + ∠ B = 90 °
ν 5x = 90, x = 18
∴∠B=36°

It is known that a, B, C are the three sides of the triangle ABC, and the square of a plus the square of B plus the square of C minus AB minus BC minus AC is equal to 0?

a^2+b^2+c^2-ab-bc-ac=0
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0
(a-b)^2+(b-c)^2+(c-a)^2=0
So (a-b) ^ 2 = (B-C) ^ 2 = (C-A) ^ 2 = 0
So A-B = B-C = C-A = 0
So a = b = C
So it's an equilateral triangle

In the triangle ABC, if the square of a plus AB equals the square of C minus the square of B, then the inner angle c is equal to?

0

0

Cosine theorem C ^ 2 = a ^ 2 + B ^ 2-2bccosc
According to the corresponding conditions, COSC = - 1 / 2, i.e. C = 120 ° can be obtained