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If x>0, y<0 and z<0, (|x|+|y|+|z|)^2=?

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If x>0, y<0 and z<0, (|x|+|y|+|z|)^2=?

by Max@Math Revolution » Sat Jan 23, 2016 10:30 pm
If x>0, y<0 and z<0, (|x|+|y|+|z|)^2=?

A. x^2+y^2+z^2+2xy+2yz+2zx
B. x^2+y^2+z^2+2xy-2yz+2zx
C. x^2+y^2+z^2-2xy+2yz-2zx
D. x^2+y^2+z^2-2xy-2yz-2zx
E. x^2-y^2-z^2+2xy+2yz+2zx


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by Max@Math Revolution » Wed Jan 27, 2016 5:27 pm
If x>0, y<0 and z<0, (|x|+|y|+|z|)^2=?

A. x^2+y^2+z^2+2xy+2yz+2zx
B. x^2+y^2+z^2+2xy-2yz+2zx
C. x^2+y^2+z^2-2xy+2yz-2zx
D. x^2+y^2+z^2-2xy-2yz-2zx
E. x^2-y^2-z^2+2xy+2yz+2zx

==> |A|=A when A>0 ,and |A|=-A when A<0.
So, (|x|+|y|+|z|)^2=(x-y-z)^2=x^2+(-y)^2+(-z)^2+2x(-y)+2(-y)(-z)+2(-z)x
=x^2+y^2+z^2-2xy+2yz-2zx.
Therefore, the answer is C.
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