Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{168}-1\right)\)\(< \)\(3^{168}-1\)
\(\Rightarrow\)\(A< B\)
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)
Vậy A<B
b, \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=A\)
Vậy A>B
\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)
\(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)
\(< 2016^2=B\)
Nên A<B
\(B=2016^2\)
\(\Rightarrow B=\left(2017-1\right)^2\)
\(\Rightarrow B=2017^2-4034+1=2017^2-4033\)(1)
Lại Có :
\(A=2015.2017=\left(2017-2\right).2017\)
\(\Rightarrow A=2017^2-4034\)(2)
Từ (1) và (2) => B>A
\(A^2=3940+2\cdot\sqrt{1970^2-1}\)
\(B^2=3940+2\cdot\sqrt{1970^2}\)
mà \(1970^2-1< 1970^2\)
nên A<B
Còn thêm cách nào khác ko ạ? Nếu có thì giúp em nha. Cảm ơn anh nhiều!
\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{128}-1\right)< B\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)
\(\Rightarrow A< B\)
\(20162017.20162019=\left(20162018-1\right)\left(20162018+1\right)\)
\(=20162018^2-1< 20162018^2\)
\(\Rightarrow20162017.20162019< 20162018^2\)
Vậy...
Ta có:
\(20162017=20162018-1\)
\(20162019=20162018+1\)
\(\Rightarrow20162017.20162019\)
\(=\left(20162018-1\right).\left(20162018+1\right)\)
\(=20162018^2-1^2=20162018^2-1\)
\(\Rightarrow20162018^2-1< 20162018^2\)
Do đó \(20162017.20162018< 20162018^2\)
Chúc bạn học tốt!!!