So sánh:\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)và\(3^{32}\)
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23 tháng 7 2021
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
10 tháng 7 2017
Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
29 tháng 6 2017
Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy
8 tháng 7 2018
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow2A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)
6 tháng 3 2020
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)
KT
7 tháng 7 2018
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(.........\)
\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)
Vậy \(B< A\)
MT
9 tháng 7 2015
A=(3+1)(32+1)(34+1)(38+1)(316+1)
=>2A=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
=>A=\(\frac{3^{32}-1}{2}
13 tháng 9 2019
Let A = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
=> 2A = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
= (38 - 1)(38 + 1)(316 + 1)(332 + 1)
= (316 - 1)(316 + 1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
⇒A=364−12
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
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