2+2+2+2+2+2+2
Ok?
Những câu hỏi liên quan
CMR:
frac{1}{2}-frac{2}{2^2}-frac{3}{2^3}-...-frac{100}{2^{100}}frac{2}{9}
Ghi cách giải giùm mk gấp nha!!!
Đọc tiếp
CMR:
\(\frac{1}{2}\)-\(\frac{2}{2^2}\)-\(\frac{3}{2^3}\)-...-\(\frac{100}{2^{100}}\)<\(\frac{2}{9}\)
Ghi cách giải giùm mk gấp nha!!!
2^x = 4 + 2^2 + 2^3 + 2^4 + ... + 2^99 + 2^100
Tìm x ( chi tiết ok ? )
Tính D, biết
D = 1+2+2 mũ 2 + 2 mũ 3 + ...+2 mũ 63
cố gắng giúp mình nhé!
\(D=1+2+2^2+2^3+.....+2^{63}\)
\(\Leftrightarrow2D=2^1+2^2+2^3+.....+2^{63}+2^{64}\)
\(\Leftrightarrow2D-D=\left(2^1+2^2+2^3+.....+2^{63}+2^{64}\right)-\left(2^0+2^1+2^2+2^3+.....+2^{63}\right)\)
\(\Leftrightarrow D=2^{64}-1\)
Vậy.............
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D = 1+2+2 mũ 2 + 2 mũ 3 + ...+2 mũ 63
2D = 2+2 mũ 2 + 2 mũ 3 + 2 mũ 4 +...+2 mũ 64
2D - D =( 2+2 mũ 2 + 2 mũ 3 + 2 mũ 4 +...+2 mũ 64) - (1+2+2 mũ 2 + 2 mũ 3 + ...+2 mũ 63)
D = 2 mũ 64 -1
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\(D=1+2+2^2+2^3+...+2^{63}\)
=>\(2D=2.\left(1+2+2^2+2^3+...+2^{63}\right)\)
=>\(2D=2+2^2+2^3+...+2^{64}\)
=>\(2D-D=\left(2+2^2+2^3+...+2^{64}\right)-\left(1+2+2^2+2^3+...+2^{63}\right)\)
=>\(D=2^{64}-1\)
Vậy \(D=2^{64}-1\)
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Chứng minh:
B=1/2+2/2^2+....+100/2^100<2
AI GIẢI MÌNH TICK OK HA
\(2B=2\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{100}{2^{100}}\right)\)
\(2B=1+1+\frac{3}{2}+...+\frac{100}{2^{99}}\)
\(2B-B=\left(2+\frac{3}{2}+...+\frac{100}{2^{99}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{100}{2^{100}}\right)\)
\(B=2-\frac{1}{2}+\frac{2}{2^2}+\frac{100}{2^{100}}<2\)
=>B<2(đpcm)
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tim x biet :
a) ( 2x2+4 ) - (x2-3/2 ) = (-3+4x2)+ ( -4x2/3+1)
b) ( I xI -4) - ( 2- I-xI ) = 1/3 . I xI -5
a: \(\Leftrightarrow2x^2+4-x^2+\dfrac{3}{2}=-3+4x^2-\dfrac{4}{3}x^2+1\)
\(\Leftrightarrow x^2+\dfrac{11}{2}=\dfrac{8}{3}x^2-2\)
\(\Leftrightarrow x^2\cdot\dfrac{-5}{3}=-\dfrac{15}{2}\)
\(\Leftrightarrow x^2=\dfrac{9}{2}\)
hay \(x\in\left\{\dfrac{3\sqrt{2}}{2};-\dfrac{3\sqrt{2}}{2}\right\}\)
b: \(\Leftrightarrow\left|x\right|-4-2+\left|x\right|-\dfrac{1}{3}\left|x\right|+5=0\)
\(\Leftrightarrow\left|x\right|\cdot\dfrac{5}{3}=1\)
hay \(x\in\left\{\dfrac{3}{5};-\dfrac{3}{5}\right\}\)
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A = 4+\(2^2+2^3+2^4+.......+2^{20}\)
Hỏi A có chia hết cho 128 ko ?
giúp mình với nha 
A = 4+2^2+2^3+2^4+.......+2^20
A=2^2+2^2+2^3+2^4+.......+2^20
2A=2.(4+2^2+2^3+2^4+.......+2^20)
2A=2^3+2^3+2^4+2^5+...+2^21
2A-A=(2^3+2^3+2^4+2^5+...+2^21-(2^2+2^3+2^4+.......+2^20)
A=2^3+2^21-(2^2+2^2)
A=8+2^21-8
A=2^21
-Ta có:128=2^7
Mà 2^21 chia hết cho 2^7
Vậy A chia hết cho 128.
Nhớ tick nha!
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A=4+22+23+24+...+220
=>2.A=23+24+25+...+221
=>2A-A=23+24+25+...+221-(22)-(22)-(23)-(24)-...-(220)
=>A= -(22)-(22)
=>A= -8
Vậy A= -8
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A=4+2\(^2\)+23+24+...+220
=>2A=2\(^3\)+2\(^3\)+24+25+...+221
=>2A-A=2\(^3\)+2\(^3\)+24+25+...+221-(22)-(2\(^2\))-(23)-(24)-...-(220)
=>A=2\(^3\)+2\(^{21}\)-(22)-(2\(^2\))
=>A=8+221-8
=>A=27.214 chia hết cho 128
Vậy A chia hết cho 128
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Cho A =\(\dfrac{1}{^{ }2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}\)
Chứng minh rằng A < 1
Ai làm nhanh mà đúng thì Đu Đủ tick nà!
Giúp Đu Đủ Thiện Nhân nha!!!!!
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(=1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1\) ( đpcm )
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mình gợi ý nè :
Chứng minh A <\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
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A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2012^2}\) + \(\dfrac{1}{2013^2}\)
Vì \(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\)
...
\(\dfrac{1}{2012^2}\) < \(\dfrac{1}{2011.2012}\)
\(\dfrac{1}{2013^2}\) < \(\dfrac{1}{2012.2013}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2012^2}\) + \(\dfrac{1}{2013^2}\) < \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{2011.2012}\) +
+ \(\dfrac{1}{2012.2013}\)
Hay A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
A < \(1-\dfrac{1}{2013}\)
A < \(\dfrac{2012}{2013}\)
Mà \(\dfrac{2012}{2013}\) < 1
\(\Rightarrow\) A < \(\dfrac{2012}{2013}\) < 1
Hay A < 1
Vậy A < 1
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tính 2 cách
\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
giúp mk nhé
\(=\frac{\frac{5}{11.2}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{1}{11}+\frac{3}{2}}=\frac{5}{\frac{2}{4}}=\frac{5}{\frac{1}{2}}\)
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6a1 ok
thank nha
mình cần gấp
Cho a=2+2^2+2^3+...+2^60
CMR a chia hết 3,7,42
A=2+22+23+.....+260
A=(2+22)+(23+24)+.......+(259+260)
A=(2+22)+22.(2+22)+......+258.(2+22)
A=6+22+6+.......+258.6
A=6.(1+22+......+258)
A=2.3.(1+22+.....+258) \(⋮\)3
Vậy A\(⋮\)3
A=2+22+23+....+260
A=(2+22+23)+.....+(258+259+260)
A=(2+22+23)+.....+257.(2+22+23)
A=14+....+257.14
A=14.(1+...+257)
A=2.7.(1+....+257)\(⋮\)7
Vậy A\(⋮\)7
A=2+22+23+....+260
A=(2+22+23+24+25+26)+....+(255+256+257+258+259+260)
A=(2+22+23+24+25+26)+.....+254.(2+22+23+24+25+26)
A=126+...+254.126
A=126.(1+....+254)
A=42.3.(1+...+254) \(⋮\)42
Vậy A\(⋮\)42
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