1/2+1/4+1/8+1/6+1/32
1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32
Tính
1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744
1*2*4+2*4*8+4*8*16+8*16*32/1*3*4+2*6*8+4*12*16+8*24*32 = 56744
#nhớ tk
1/2+1/4+1/8+1/6+1/32
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(A=1-\frac{1}{32}\)
\(A=\frac{31}{32}\)
{ 1/2 + 1/4 + 1/8 + 1/16 } : x = 1/2 + 1/6 + 1/8 + 1/12 + ....+ 1/32
Bài 1 Tính nhanh
a) -3/7 + 5/13 + 3/7
b) -5/21+-2/21+8/24
c) -5/11+(-6/11+2)
d) (-1/32+1/2)+15/32
e)5/17+ -6/13 + 3/4 + 7/-13+12/17
f) 7/23+-18/18+-4/9+16/23+-5/8
g)1/3+-3/4+3/5+-1/36+1/15+-2/9
h)-1/2+1/3+-1/4+-2/8+4/18+4/9
a)\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{3}{7}\)
=\(\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\dfrac{5}{13}\)
=\(0+\dfrac{5}{13}\)
=\(\dfrac{5}{13}\)
{ 1/2 + 1/4 + 1/8 + 1/16 } : x = 1/2 + 1/6 + 1/12 + .....+1/32
1/4*2/6*3/8*…*14/30*15/32=1/2^2x+1
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
B=2*4+2*4*8+4*8*16+8*16*32 / 3*4+2*6*8+4*12*16+8*24*32 các bạn có thể giả bằng 1 số nhân với 1 tổng ko ?
bài 6:tính nhanh
7)1\(^2\)-2\(^2\)+3\(^2\)-4\(^2\)+....-2004\(^2\)+2005\(^2\)
8) (2+1)(2\(^2\)+1)(2\(^4\)+1)(2\(^8\)+1)(2\(^{16}\)+1)(2\(^{32}\)+1)-2\(^{64}\)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
Chứng minh rằng 1/2 - cho 1/4 + 1/8 - cho 1/16 + 1/32 - 1/6 4 <1/3