Bài 6. So sánh hai số bằng cách vận dụng hằng đẳng thức:
a. \(A=1999.2001\)và \(B=2000^2\)
b. \(A=2^{16}\)và \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
c. \(A=2011.2013\)và \(B=2012^2\)
d. \(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\) và \(B=3^{128}-1\)
Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)
\(< =>A>B\)
a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)
\(\Rightarrow A< B\)
b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
...
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)
\(\Rightarrow A>B\)
c,d tương tự
Ta có
\(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)
Mà \(B=2012^2\)
Suy ra \(B>A\)
d,\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(.....\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(=3^{128}-1\)
\(\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1\)
\(\Rightarrow A< B\)
