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\(a.\)
Ta sẽ biến đổi biểu thức \(B\) quy về dạng có thể dùng được hằng đẳng thức \(\left(x-y\right)\left(x+y\right)=x^2-y^2\), khi đó:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
Vì \(2^{16}>2^{26}-1\) nên \(2^{16}>\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
Vậy, \(A>B\)
Tương tự với câu \(b\) kết hợp với phương pháp tách hạng tử, khi đó xuất hiện hằng đẳng thức mới và dễ dàng đơn giản hóa biểu thức \(A\). Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\frac{1}{2}\left(3^{128}-1\right)\)
Mặt khác, do \(\frac{1}{2}<1\) nên \(\frac{1}{2}\left(3^{128}-1\right)<3^{128}-1\)
Vậy, \(B>A\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{22}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)
Chúc bạn học tốt !!!
A.(32-1)=4.(32-1)(32+1)(34+1)...(364+1)=4.(34-1)(34+1)...(364+1)= ... =4.(3128-1)
<=>8A=4B <=>2A=B =>B>A
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
\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{128}-1\right)< B\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)
\(\Rightarrow A< B\)
\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\dfrac{1}{2}\left(3^{128}-1\right)< 3^{128}-1=Q\)
\(P=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\\ 2P=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\\ 2P=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1\\ P=\dfrac{3^{128}-1}{2}< Q=3^{218}-1\)