e: \(2\sqrt{26}>9\)

nên \(2\sqrt{26}+4>13\)

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 \(10A< 10B\) 

\(\Rightarrow A< B\)

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 

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: \(\dfrac{25}{24}>\dfrac{36}{37}\)

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

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

a, Chia hết cho 3 thì nhóm 2 số thành 1 cặp ; chia hết cho 7 thì nhóm 3 số thành 1 cặp

b, Đề phải là A = 2009.2011

Có :A = 2009.(2010+1) = 2009.2010+2009

= 2009.2010+2010-1 = 2010.(2009+1)-1 = 2010^2-1

Vì 2010^2-1 < 2010^2 = B => A < B

c, A = (3^3)^150 = 27^150

B = (5^2)^150 = 25^150

Vì 27^150 > 25^150 => A > B

k mk nha

bai nao vay