tính 1-1/2-1/4-1/8-1/16 giúp tôi với
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b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
\(A=\left(a\text{x}7+a\text{x}8-a\text{x}15\right):\left(1+2+3+...+10\right)\)
\(A=\left(a\text{x}\left(7+8-15\right)\right):\left(1+2+3+...+10\right)\)
\(A=\left(a\text{x}0\right):\left(1+2+3+..+10\right)\)
\(A=0:\left(1+2+3+...+10\right)\)
\(A=0\)
\(B=\left(18-9\text{x}2\right)\text{x}\left(2+4+6+8+10\right)\)
\(B=\left(18-18\right)\text{x}\left(2+4+6+8+10\right)\)
\(B=0\text{x}\left(2+4+6+8+10\right)\)
\(B=0\)
dưới mẫu nè: (2+1)(2^2+1)(2*4+1)(2*8+1)(2*16+1)=(2*4-1)(2*4+1)(2*8+1)(2*16+1)(*vì 2+1=2*2-1)
cứ như thế thì được: 2*32-1
Ta có : \(\frac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\frac{\left(2^4\right)^8-1}{\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\frac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\frac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\frac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\frac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\frac{2^{32}-1}{2^{32}-1}=1\)
Ta có:2A=\(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
2A-A=\(\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(=2-\frac{1}{32}=\frac{63}{32}=A\)
Ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\right)\)
\(\Rightarrow A=1-\frac{1}{2^5}=\frac{31}{32}\)
Vậy \(A=\frac{31}{32}\)
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)
=>\(B=\dfrac{32+16+6+2+1}{64}\)
=>\(B=\dfrac{63}{64}\)
Ta có : ( 2+1 ) ( 22 +1 ) (24 +1 ) ( 28 +1 ) ( 216 +1 )
= (22 - 1)(22 + 1)(24 +1 ) ( 28 +1 ) ( 216 +1 )
= (24 - 1)(24 +1 ) ( 28 +1 ) ( 216 +1 )
= (28 - 1) ( 28 +1 ) ( 216 +1 )
= (216 - 1 ) ( 216 +1 )
= 232 - 1 (đpcm)
1/16 NHA
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