[tex]\displaystyle \sf \text{Sabemos que }: \\\\ a^{(p+q)} = a^{p}\cdot a^{q} \\\\\ \text{Da{\'i}, temos} : \\\\ 2^{51}-2^{50}-2^{49} \\\\ 2^{49}\cdot 2^{2}-2^{1}\cdot 2^{49}-2^{49}\\\\ 4\cdot 2^{49}-2\cdot 2^{49}-2^{49} \\\\\ \underline{\text{Colocando o\ }2^{49} \text{em evid\^encia}}: \\\\ 2^{49}\cdot (4-2-1) \\\\ 2^{49}\cdot 1 \\\\ Portanto : \\\\\ \Large\boxed{\sf 2^{51} - 2^{50} - 2^{49} = 2^{49}\ }\checkmark[/tex]
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[tex]\displaystyle \sf \text{Sabemos que }: \\\\ a^{(p+q)} = a^{p}\cdot a^{q} \\\\\ \text{Da{\'i}, temos} : \\\\ 2^{51}-2^{50}-2^{49} \\\\ 2^{49}\cdot 2^{2}-2^{1}\cdot 2^{49}-2^{49}\\\\ 4\cdot 2^{49}-2\cdot 2^{49}-2^{49} \\\\\ \underline{\text{Colocando o\ }2^{49} \text{em evid\^encia}}: \\\\ 2^{49}\cdot (4-2-1) \\\\ 2^{49}\cdot 1 \\\\ Portanto : \\\\\ \Large\boxed{\sf 2^{51} - 2^{50} - 2^{49} = 2^{49}\ }\checkmark[/tex]
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