1/2+1/4+1/8+1/6+1/32
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B
4
19 tháng 6 2018
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}\)
\(A=\frac{31}{32}\)
LV
0
LV
0
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 7 2023
ta có: 4 = 2 x 2
6 = 2 x 3
...... = ..........
30 = 2 x 15
Nhân vế với vế ta có: 4x6x8x...x30 = 214x (2x3x4x...x15)
⇒ \(\dfrac{1}{4}\times\dfrac{2}{6}\times\dfrac{3}{8}\times...\times\dfrac{14}{30}\times\dfrac{15}{32}\) = \(\dfrac{2\times3\times...\times14\times15}{2^{14}\times\left(2\times3\times...\times14\times15\right)\times32}\)
⇒ \(\dfrac{1}{2^{14}\times2^5}\) = \(\dfrac{1}{2^{2x+1}}\) ⇒ 219 = 2\(2x\)+1
⇒ 19 = 2\(x\) + 1 ⇒ 2\(x\) = 18 \(\Rightarrow\) \(x\) = 9
ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
NL
0
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{32}=\dfrac{48+24+12+16+3}{96}=\dfrac{103}{96}.\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{32}=\dfrac{1.48}{2.48}+\dfrac{1.24}{4.24}+\dfrac{1.12}{8.12}+\dfrac{1.16}{6.16}+\dfrac{1.3}{32.3}=\dfrac{48+24+12+16+3}{96}=\dfrac{103}{96}\)