3條math ........幫下我

(1)

　(x-y+z)²
=[(x-y)+z]²
=(x-y)²+2(x-y)(z)+z²
=x²-2xy+y²+2xz-2yz+z²
=x²+y²+z²-2xy+2xz-2yz

(2)

　50-8y²
=2(25-4y²)
=2[(5²-(2y)²]
=2(5-2y)(5+2y)

(3)
The first cicrle circimference=2 π r1
The second circle circumference=2 π r2

a)the difference of the square of their circumference
=(2 π r1)²-(2 π r2)²
=(2 π r1 - 2 π r2)(2 π r1 + 2 π r2)
=2π(r1-r2)2π(r1+r2)
=4π²(r1-r2)(r1+r2)

b)the square of the difference
=(2 π r1 - 2 π r2)²
=[2π(r1 - r2)]²
=4π²(r1 - r2)²

c)the sum of the square
=(2 π r1)² + (2 π r2)²
=4 π² r1² + 4 π r2²
=4π²(r1² + r2²)

1) (x-y+z)^2
=x(x-y+z)-y(x-y+z)+z(x-y+z)
=X^2-y^2+z^2-2xy+2xz-2yz

2)50-8y^2
=2(25-4y^2)
=2(5^2-2^y^2)
=2(5^2-(2y)^2)
=2(5+2y)(5-2y)

1)
(x-y+z)*(x-y+z)
=x^2 -xy + xz -xy +y^2 - yz +xz -yz +z^2
=x^2 + y^2 + z^2 -2xy +2xz -2yz

1. x^2+y^2+z^2-2xy-2yz
2.= 2(25-4y^2)
=2(5-2y)(5+2y)
3.
a) r1^2 - r2^2
b) (r1-r2)^2
c) (2πr1)^2 + (2πr2)^2 = 4π^2 (r1^2+r2^2)