A=1+2+2^2+2^3+....+2^9

2A=2+2^2+2^3+....+2^10

2A-A=2^10-1

A=2^10-1/2

B=5.2^8=(2^2+1).2^8=2^10+2^8

=>B>A

2A = 2(1 + 2 + 2² + .... + 2⁹ )

= 2 + 2² + 2³ + ..... + 2¹⁰

2A - A = (2 + 2² + 2³ + ..... + 2¹⁰) - (1 + 2 + 2² + .... + 2⁹ )

A = 2¹⁰ - 1  

B = 5.2⁸ = (2² + 1).2⁸ = 2¹⁰ + 2⁸

2¹⁰ - 1 < 2¹⁰ + 2⁸

=> A < B

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

Bài 1: 

a: \(-2^{30}=-8^{10}\)

\(-3^{30}=-27^{10}\)

mà 8<27

nên \(-2^{30}>-3^{30}\)

b: \(35^5=35^5\)

\(6^{10}=36^5\)

mà 35<36

nên \(35^5< 6^{10}\)

a)

Có: 

\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)

Vì \(\sqrt{117}>\sqrt{116}\)  nên \(3\sqrt{13}>2\sqrt{29}\)

b)

Có:

\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)

\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)

Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\)  nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)

c)

Có:

\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)

\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)

Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)

d)

Có:

\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)

\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)

lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)

 \(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)

Giải:

a) Gọi dãy đó là A, ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\) 

\(2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\) 

\(2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\right)\) 

\(A=\dfrac{1}{2}-\dfrac{1}{2^{2014}}\) 

Vì \(\dfrac{1}{2}< 1;\dfrac{1}{2^{2014}}< 1\) nên \(\dfrac{1}{2}-\dfrac{1}{2^{2014}}< 1\) 

\(\Rightarrow A< 1\) 

b) \(A=\dfrac{10^{11}-1}{10^{12}-1}\) và \(B=\dfrac{10^{10}+1}{10^{11}+1}\) 

Ta có:

\(A=\dfrac{10^{11}-1}{10^{12}-1}\) 

\(10A=\dfrac{10^{12}-10}{10^{12}-1}\) 

\(10A=\dfrac{10^{12}-1+9}{10^{12}-1}\) 

\(10A=1+\dfrac{9}{10^{12}-1}\) 

Tương tự:

\(B=\dfrac{10^{10}+1}{10^{11}+1}\) 

\(10B=\dfrac{10^{11}+10}{10^{11}+1}\) 

\(10B=\dfrac{10^{11}+1+9}{10^{11}+1}\) 

\(10B=1+\dfrac{9}{10^{11}+1}\) 

Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\) nên \(10A< 10B\) 

\(\Rightarrow A< B\)

a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)

b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)

c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)

d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)

1:

a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)

b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)

nên 2 p/s ko bằng nhau