Evaluate:
80+40+208^0 + 4^0 + 2^080+40+20
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According to property of exponents, a0=1a^0 = 1a0=1
80+40+20=1+1+1=38^0 + 4^0 + 2^0 \\[1em] = 1 + 1 + 1 \\[1em] = 380+40+20=1+1+1=3
Hence, 80+40+20=38^0 + 4^0 + 2^0 = 380+40+20=3
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Simplify:
(2−3−2−4)(2−3+2−4)(2^{-3} - 2^{-4})(2^{-3} + 2^{-4})(2−3−2−4)(2−3+2−4)
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(8+4+2)0(8 + 4 + 2)^0(8+4+2)0
4x04x^04x0