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Áp dụng liên tục a2 - b2 = (a - b)(a + b) để biến đổi . Ta có:
A = 332 - 1 = (316 - 1)(316 + 1) = (38- 1)(38 + 1)(316 + 1) = (34 - 1)(34 + 1)(38 + 1)(316 + 2) = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) =
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) = 2.B
Ta có 2B = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
2B = (34-1)(34+1)(38+1)(316+1)
2B = (38-1)(38+1)(316+1)
Tương tự ta đc:
2B = 332-1
B= 332-1/2 hay B= A/2
Vậy A>B
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)
Vậy \(B< A\)
Ta có: B=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow\) 2B= \(2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
= \(\left(3^{16}-1\right)\left(3^{16}+1\right)\)
= \(3^{32}-1\)
\(\Rightarrow\) B= \(\dfrac{3^{32}-1}{2}\)
Mà ta có A= \(3^{32}-1\)
\(\Rightarrow\) A=2B
Nếu đề thế này thì mình có thể làm được:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\dfrac{3^{32}-1}{2}\)
=> B>A
Ta có: \(A=3^{32}-1=\left(3^{16}+1\right)\left(3^{16}-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^8-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(=\left(3^{16}+1\right)\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Vậy A = 2B
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)
= (38 - 1)(38 + 1)(316 + 1)
= (316 - 1)(316 + 1)
= 332 - 1 < 332
Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :
\(y^2+2y+4^x-2^{x+1}+2=0\)
\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)
\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)
\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)
CHÚC BẠN HỌC TỐT...........
Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)