Y = root sign (x - 2) + follow sign (4 - x) for range

Y = root sign (x - 2) + follow sign (4 - x) for range

The domain of y = √ (X-2) + √ (4-x) is [2,4]. In the domain of Y > 0y ^ 2 = (X-2) + 2 √ [(X-2) (4-x)] + (4-x) = 2 + 2 √ [- x ^ 2 + 6x-8] u = - x ^ 2 + 6x-8, the axis of symmetry is x = 3, so in the interval [2,4], u gets the maximum value of 1 at x = 3, and the minimum value of 0 at x = 2 and x = 4. The maximum value of Y ^ 2 is 2 + 2 = 4

Given a + B = - 5, ab = 7, find the value of a 2B + ab 2-a-b

∵a+b=-5，ab=7，∴a2b+ab2-a-b=ab（a+b）-（a+b）=（ab-1）（a+b）=（7-1）（-5）=-30．

If a + B = 7, ab = 5, then the value of (a-b) & sup2

（a-b）²
=a^2+b^2-2ab
=a^2+b^2+2ab-4ab
=(a+b)^2-4ab
=49-20
=29

Given: a + B + C = 0, a & sup2; + B & sup2; + C & sup2; = 1, find the values of AB + BC + Ca and a ^ 4 + B ^ + C ^ 4

Given a + B + C = 1, a 2 + B 2 + C 2 = 2, find the value of AB + BC + ca

∵ a + B + C = 1, ∵ a + B + C) 2 = A2 + B2 + C2 + 2Ab + 2BC + 2Ac = 1, ∵ A2 + B2 + C2 = 2, ∵ 2 + 2Ab + 2BC + 2Ac = 1, the solution is ab + BC + AC = - 12

Given A-B = 3, B-C = - 2, find the value of a & sup2; + B & sup2; + c-ab-bc-ca

a-b=3,b-c=-2∴a-b+b-c=a-c=1a²+b²+c-ab-bc-ca=½[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]=½[(a-b)²+(a-c)²+(b-c)²]=½×﹙3²＋2²＋...

It is known that A.B.C is the three sides of △ ABC, and satisfies a & sup2; + B & sup2; + C & sup2; - AB BC CA = 0. Try to judge what kind of special triangle △ ABC is

a²+b²+c²-ab-bc-ca=0
That is 2A & sup2; + 2B & sup2; + 2C & sup2; - 2ab-2bc-2ca = 0 (both sides of the equation multiply by 2)
a²+a²+b²+b²+c²+c²-2ab-2bc-2ca=0
(a²+b²-2ab)+(b²+c²-2bc)+(a²+c²-2ac)=0
(a-b)²+(b-c)²+(a-c)²=0
And ∵ (a-b) & sup2; ≥ 0, (B-C) & sup2; ≥ 0, (A-C) & sup2; ≥ 0
∴a-b=0,b-c=0,a-c=0
The solution is a = B, B = C, a = C
∴a=b=c
That is, ABC is an equilateral triangle

If a & sup2; + B & sup2; + C & sup2; = AB + BC + Ca, prove a = b = C
As the title

A & sup2; + B & sup2; + C & sup2; = AB + BC + Ca2 (A & sup2; + B & sup2; + C & sup2;) = 2 (AB + BC + Ca) a & sup2; - 2Ab + B & sup2; + A & sup2; - 2Ac + C & sup2; + B & sup2; - 2BC + C & sup2; = 0 (a-b) & sup2; + (A-C) & sup2; + (B-C) & sup2; = 0A = b = C, didn't I tell you

If a + 2B + 3C = 12 and a & sup2; + B & sup2; + C & sup2; = AB + BC + Ca, then a + B & sup2; + C & sup3=_____ .
Recognition of "multiplication consensus" -- factorization (2)

14
a^2+b^2+c^2=ab+bc+ca
2a^2+2b^2+2c^2=2ab+2bc+2ca
(a-b)^2+(c-a)^2+(b-c)^2=0
a=b=c; a+2b+3c=12
a=b=c=2,a+b^2+c^3=14.

Calculation: (2b-3c + 4) (3c-2b + 4) - 2 (B-C) 2=___ ．