Question

Trong hai số sau số nào lớn hơn:

C=\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)và D=\(2^{16}\)

Answer 1

Lời giải:

Sử dụng công thức: \((a-b)(a+b)=a^2-b^2\) ta có:

\(C=(2+1)(2^2+1)(2^4+1)(2^8+1)\)

\(\Leftrightarrow C=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)\)

\(\Leftrightarrow C=(2^2-1)(2^2+1)(2^4+1)(2^8+1)\)

\(\Leftrightarrow C=(2^4-1)(2^4+1)(2^8+1)\)

\(\Leftrightarrow C=(2^8-1)(2^8+1)=2^{16}-1<2^{16}\)

Do đó \(C< D\)

Answer 2

\(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(C=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(C=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(C=\left(2^8-1\right)\left(2^8+1\right)\)

\(C=2^{16}-1< 2^{16}\)