\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

A=20+21+22+23+...++23+...+250250

2�=2+22+23+...+2512A=2+22+23+...+251

2�−�=�=251−202A−A=A=251−20

�=5+52+53+...+599+5100B=5+52+53+...+599+5100

5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101

5�−�=4�=5101−55B−B=4B=5101−5

�=5101−54B=45101−5​

�=3−32+33−34+...+C=3−32+33−34+...+32007−32008+32009−3201032007−32008+32009−32010

3�=32−33+34−35+...−32008+32009−32010+320113C=32−33+34−35+...−32008+32009−32010+32011

3�+�=4�=32011+33C+C=4C=32011+3

�=32011+34C=432011+3​

�100=5+5×9+5×92+5×93+...+5×999S100​=5+5×9+5×92+5×93+...+5×999

�100=5×(1+9+92+93+...+999)S100​=5×(1+9+92+93+...+999)

9�100=5×(9+92+93+...+999+9100)9S100​=5×(9+92+93+...+999+9100)

9�100−�100=8�100=5×(9100−1)9S100​−S100​=8S100​=5×(9100−1)

�100=5×(9100−1)8S100​=85×(9100−1)​

Bài 1

a) A = 2^0 + 2^1 + 2^2 +...+ 2^50

2A=2^1+2^2+2^3+...+2^51

2A-A=(2^1+2^2+2^3+...+2^51)-(2^0 + 2^1 + 2^2 +...+ 2^50)

A=(2^1-2^1)+(2^2-2^2)+...+(2^50-2^50)+(2^51-2^1)

A=0+0+...+0+(2^51-2^1)

A=2^51-2^1

b)B = 5 + 5^2 + 5^3 +...+ 5^99 + 5^100

5B=5^2+5^3+5^4+...+5^100+5^101

5B-B=(5^2+5^3+5^4+...+5^100+5^101)-( 5 + 5^2 + 5^3 +...+ 5^99 + 5^100)

4B=(5^2-5^2)+(5^3-5^3)+...+(5^100-5^100)+(5^101-5)

4B=0+0+...+0+(5^101-5)

4B=5^101-5

B=(5^101-5)/4

c)C = 3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010

3C=3^2-3^3+3^4-3^5+...+3^2010-3^2011

3C-C=(3^2-3^3+3^4-3^5+...+3^2010-3^2011)-(3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010)

...............................................!!!!!!!!!!!!!!!!!!!!!!!!

Bài 2

8(mình k0 chắc)

Làm bài 1 cũng đc rồi. Cảm ơn bạn nhiều

2)x=10

A=-2/3

B=1

a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)

A= 1+5+5^2+5^3+...+5^51

=> 5A= 5+5^2+5^3+5^4+...+5^52

=> 5A - A= ( 5+5^2+5^3+5^4+...+5^52) -(1+5+5^2+5^3+...+5^51)

=> 4A = 5^52-1

=>A=(5^52-1)/4

a) 2√18 - 4√50 + 3√32

= 6√2 - 20√2 + 12√2

= -2√2

b) √(√8 - 4)² + √8

= 4 - √8 + √8

= 4

c) √(14 - 6√5) + √(6 + 2√5)

= √(3 - √5)² + √(√5 + 1)²

= 3 - √5 + √5 + 1

= 4

\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)

a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)