\(a,2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\\ \Leftrightarrow6+2\sqrt{2}< 3+6=9\\ b,\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}\\ 2^2=4=14-10\\ \left(2\sqrt{33}\right)^2=132>100=10^2\Leftrightarrow-2\sqrt{33}< -10\\ \Leftrightarrow\sqrt{11}-\sqrt{3}< 2\)

a: \(2\sqrt{2}< 3\)

nên \(6+2\sqrt{2}< 9\)

a) `14/21=(14:7)/(21:7)=2/3=4/6`

`60/72=(60:12)/(72:12)=5/6`

Vì `4/6 <5/6`

`=> 14/21 < 60/72`

b) `22/37 = (22:2)/(37:2)= 11/(37/2)`

Vì `54 > 37/2`

`=> 11/54 < 22/37` 

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 \(10A< 10B\) 

\(\Rightarrow A< B\)

Bài 2: 

a: Ta có: \(\dfrac{9}{11}=1-\dfrac{2}{11}\)

\(\dfrac{13}{15}=1-\dfrac{2}{15}\)

mà \(-\dfrac{2}{11}< -\dfrac{2}{15}\)

nên \(\dfrac{9}{11}< \dfrac{13}{15}\)

b: Ta có: \(\dfrac{19}{15}=1+\dfrac{4}{15}\)

\(\dfrac{15}{11}=1+\dfrac{4}{11}\)

mà \(\dfrac{4}{15}< \dfrac{4}{11}\)

nên \(\dfrac{19}{15}< \dfrac{15}{11}\)

2 câu kia đâu rùi

Bài 2: 

a: Ta có: 1315=1−2151315=1−215

mà 911<1315911<1315

b: Ta có: 1511=1+4111511=1+411

mà 1915<1511

\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)

Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)

b,\(3^{500}=\left(3^5\right)^{100}=243^{100}\)

Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)

cứu

a: 16^19=(2^4)^19=2^76

8^25=(2^3)^25=2^75

mà 76>75

nên 16^19>8^25

b: 3^500=(3^5)^100=243^100>5^100

\(a) 3^{200}=(3^2)^{100}=9^{100}\\2^{300}=(2^3)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\) nên \(3^{200}>2^{300}\)

\(b) 5^{40}=(5^4)^{10}=625^{10}\\3^{50}=(3^5)^{10}=243^{10}\)

Vì \(625^{10}>243^{10}\) nên \(5^{40}>3^{50}\)

#\(Toru\)

a> \(3^{200}\) và \(2^{300}\)

Ta có:\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

          \(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

Vì 9>8 nên \(9^{100}>8^{100}\)

\(\Rightarrow\)\(3^{200}>2^{300}\)

b> \(5^{40}\) và \(3^{50}\)

Ta có:\(5^{40}=5^{4.10}=\left(5^4\right)^{10}=625^{10}\)

         \(3^{50}=3^{5.10}=\left(3^5\right)^{10}=243^{10}\)

Vì 625 > 243 nên \(625^{10}>243^{10}\)

\(\Rightarrow\)\(5^{40}>3^{50}\)

`3^200=(3^2)^100=9^100`.

`2^300=(2^3)^100=8^100`.

`=> 2^300 < 3^200`.

`b, 5^40=(5^4)^10=625^10.`

`3^50=(3^5)^10=243^10`.

`=> 5^40 > 3^50`.

b)

a = 25.26 261 = 25.(26 260 +1) = 25.10.2626 + 25 = 25.10.26.101 + 25

b = 26.25 251 = 26.(25 250 + 1) = 26.10.2525 + 26 = 26.10.25.101 + 26

Suy ra a < b

a=25.26261=25.(26260+1) = 25.10.2626+25 = 25.10.26.101+25

b=26.25251=26.(25 250+1)=26.10.2525+26=26.10.25.101+26

Vì 26>25 nên b>a

a<b vì:

25<26

bài nào

bài nào bn

???