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Ta có: n – m = 3 ⇒ n = m + 3 (3)
0 < 3 ⇒ 0 + m < 3 + m ⇒ m < m + 3 (4)
Từ (3) và (4) suy ra: m < n
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Ta có: m - 1/2 = n => m - n = 1/2 => m - n > 0 => m > n.
Đáp án cần chọn là: D
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Ta có: m + 1/2 = n => m - n = - 1/2 => m - n < 0 => m < n.
Đáp án cần chọn là: A
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\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=2^{2016}-1< 2^{2016}=M\)
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\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
Ta có: m – n = 2 ⇒ m = n + 2 (1)
0 < 2 ⇒ 0 + n < 2 + n ⇒ n < n + 2 (2)
Từ (1) và (2) suy ra: n < m