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\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
Lời giải:
$22+23-25+27-29+31-33$
$=22+(23-25)+(27-29)+(31-33)$
$=22+(-2)+(-2)+(-2)=22+(-2).3=22-6=16$
So sánh : 2^33 và 3^22
2^33 = (2^3)^11 = 8^11
3^22 = (3^2)^11 9^11
Vì 8^11 < 9^11
Vậy : 2^33 < 3^22
Ta có : 2\(^{23}\)= .2\(^{20}\) . 2\(^3\) = ( 2\(^4\))\(^5\). 2\(^3\)= 16\(^5\) . 2\(^3\)
3\(^{22}\) = 3\(^{20}\) . 2\(^2\)= ( 3\(^4\))\(^5\).2\(^2\)= 81\(^5\). 2\(^2\)
Vì 16\(^5\)< 81\(^5\)nên 2\(^{23}\)< 3\(^{22}\)
\(A+2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\cdot3+...+2^{99}\cdot3\)
\(=6\left(1+...+2^{99}\right)⋮6\)
Đặt A = \(1+2+2^2+2^3+2^4+....+2^{100}\)
2A = \(2\left(1+2+2^2+2^3+2^4+....+2^{100}\right)\)
= \(2+2^2+2^3+2^4+2^5+...+2^{101}\)
2A - A = \(\left(2+2^2+2^3+2^4+2^5+....+2^{101}\right)-\left(1+2^2+2^3+2^4+...+2^{100}\right)\)
= \(2^{101}-1\)
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
\(2^{x+3}.2=2^2.3+52\)
\(=>2^{x+3}.2=64\)
\(=>2^{x+3}=64:2\)
\(=>2^{x+3}=32\)
\(=>2^{x+3}=2^5\)
=>x+3=5
=>x=5-3
=>x=2
Vậy ...........
2x + 3 . 2 = 22 . 3 + 52
2x + 3 . 2 = 4 . 3 + 52
2x + 3 . 2 = 12 + 52
2x + 3 . 2 = 64
2x + 3 = 64 : 2
2x + 3 = 32
2x + 3 = 25
x + 3 = 5
x = 5 - 3
x = 2
Vậy x = 2
Ta gọi tử của phân số B là A ta có:
A=1+2+2^2+2^3+...+2^2008
2A=2 + 2^2 + 2^3 + 2^4 +... + 2^2009
=>A=2^2009 - 1
A=-1 + 2^2009
ta thấy tử là số đối của mẫu =>B=\(\dfrac{-1}{1}\)