A= 4 x ( 3^2+1 ) x ( 3^4+1 ) x ( 3^8+1 ) x ( 3^16+1 ) và B= 3 ^32 -1
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\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)
Vậy \(A< B\)
1.a
1/2+1/4+1/8+1/16+1/32
= 1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
= 1-1/32=31/32
1b
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3} +\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}+\frac{1}{4}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{5}{20}+\frac{5}{30}+\frac{20}{30}+\frac{5}{20}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{5}{20}+\frac{5}{20}+\frac{1}{20}\right)+\left(\frac{5}{30}+\frac{20}{30}+\frac{1}{30}\right)\)
\(=\frac{11}{20}+\frac{26}{30}\)
\(=\frac{11}{20}+\frac{13}{15}\)
\(=\frac{17}{12}\)
a) A = 2016.2018 = ( 2017 - 1 ).2018 = 2017.2018 - 2018 ( 1 )
B = 20172 = 2017.2017 = 2017.( 2018 - 1) = 2017.2018 - 2017 ( 2 )
Từ (1) và (2), ta thấy: - 2018 < - 2017 => 2017.2018 - 2018 < 2017.2018 - 2017 <=> 2016.2018 < 20172
Vậy A < B
~ Phần b khi nào nghĩ ra tớ sẽ làm ngay ạ :) Còn phần này chắc chắn đúng cậu nhé ~
b)\(x=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^{16}-1\right)\left(3^{16}+1\right)\Rightarrow x=\frac{3^{32}-1}{2}\)
Thấy \(x=\frac{3^{32-1}}{2}< 3^{32}-1=y\)
1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321
=>
2.
3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2
3 - (15/8 + X - 35/24) : 4 = 2
3 - (15/8 + X - 35/24) = 2 . 4
3 - (15/8 + X - 35/24) = 8
15/8 + X - 35/24 = 3 - 8
15/8 + X - 35/24 = -5
15/8 + X = -5 + 35/24
15/8 + X = -85/24
X = -85/24 - 15/8
X = -65/12
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)
=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)
=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)
b) \(3x^2-2=72\)=> 3x2 = 74 => x2 = 74/3 => x không thỏa mãn
c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)
=> \(\left(19x+50\right):14=9\)
=> \(19x+50=126\)
=> \(19x=76\)
=> x = 4
d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)
=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)
=> \(x\left(2+4+8+16+32\right)=343\)
=> x . 62 = 343
=> x = 343/62
Có: \(A=4\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=...........................\)
\(=\frac{3^{32}-1}{2}\)
\(B=3^{32-1}\)
=> \(A< B\)