The square of a plus the square of B plus the square of C minus 2Ab minus 2BC plus 2Ac Common factor···

The square of a plus the square of B plus the square of C minus 2Ab minus 2BC plus 2Ac Common factor···

a²+b²+c²-2ab-2bc+2ac
=(a-b+c)²

a. B. C is a real number, and the square of a + the square of B + the square of C + 2a-4b + 6C + 14 is equal to 0, find the value of a.b.c

A + B + C + 2a-4b + 6C + 1 + 4 + 9 is equal to (a + 1) square + (b-2) square + (c + 3) square = 0, a = - 1, B = 2, C = - 3

Given a & # 178; + B & # 178; + C & # 178; - 2A + 4b-6c + 14 = 0, find the value of C-A + B

a²+b²+c²-2a+4b-6c+14=0(a²-2a+1)+(b²+4b+4)+(c²-6c+9)=0(a-1)²+(b+2)²+(c-3)²=0a-1=0,b+2=0,c-3=0a=1,b=-2,c=3c-a+b=3-1-2=0

In the triangle ABC, if the second power of a = the second power of B + B times the second power of C + C, then a=

120 degrees

-The 2007 power of 3 times the 2008 power of - 1 / 3

-The 2007 power of 3 times the 2008 power of - 1 / 3
=[(-3)^2007]×[(-1/3)^2008]
=[(-3)^2007]×{[1/(-3)]^2008}
=[(-3)^2007]×{1/[(-3)^2008]}
=[(-3)^2007]/[(-3)^2008]
=(-3)^(2007-2008)
=(-3)^(-1)
=1/[(-3)^1]
=-1/3
In order to make the landlord see the process clearly, it is wordy

Given X & sup2; + Y & sup2; + Z & sup2; = XY + YZ + ZX, we need to answer in detail to prove x = y = Z

Prove: ∵ X & sup2; + Y & sup2; + Z & sup2; = XY + YZ + ZX
∴x²+y²+z²-xy-yz-zx=0
Multiply both sides by two at the same time
2x²+2y²+2z²-2xy-2yz-2zx=0
That is X & sup2; - 2XY + Y & sup2; + Y & sup2; - 2yz + Z & sup2; + X & sup2; - 2zx + Z & sup2; = 0
∴(x-y)²+(y-z)²+(x-z)²=0
∵(x-y)²≥0,(y-z)²≥0,(x-z)²≥0
∴(x-y)²=0,(y-z)²=0,(x-z)²=0
∴x-y=0,y-z=0,x-z=0
∴x=y,y=z,x=z
That is, x = y = Z

X + y + Z = 2Y = 3Z (Y ≠ 0), find the value of (XY + YZ + ZX) / (X & sup2; + Y & sup2; + Z & sup2;)

Let x + y + Z = 2Y = 3Z = 6ky = 3kz = 2kx = K (XY + YZ + ZX) / (X & sup2; + Y & sup2; + Z & sup2;) = (3K & sup2; + 6K & sup2; + 2K & sup2;) / (K & sup2; + 9K & sup2; + 4K & sup2;) = (11K & sup2;) / (14K & sup2;) = 11 / 14

Given X-Y = 5, Y-Z = 2, find X & sup2; + Y & sup2; + Z & sup2; - XY YZ ZX

X-Y = 5, Y-Z = 2, add x-z = 7x & sup2; + Y & sup2; + Z & sup2; - xy-yz-xz = (2x & sup2; + 2Y & sup2; + 2Z & sup2; - 2xy-2yz-2xz) / 2 = [(X & sup2; - 2XY + Y & sup2;) + (Y & sup2; - 2yz + Z & sup2;) + (Z & sup2; - 2XZ + X & sup2;)] / 2 = [(X-Y) & sup2; + (Y-Z) & sup2; + (x-z) & sup

Let a1.a2.a3.a4 be an equal ratio sequence, and its common ratio is 2. What is the value of 2A + A2 of 2A3 + A4

a2=2a1 ,a3=4a1,a4=8a1
Then the original formula = (2A1 + 4A1) / (8A1 + 8A1) = 3 / 8

3 (a2-4a-5) · (3a-2) how to decompose a factor? What is the basis for decomposing this factor?

=3(a-5)(a+1)(3a-2)