a) 0,(26)<0,261

b) 0,15>0,14(9)

a: 0,(26)<0,261

b: 0,15>0,14(9)

a: \(\left(4+\sqrt{33}\right)^2=49+8\sqrt{33}=49+2\cdot\sqrt{528}\)

\(\left(\sqrt{29}+\sqrt{14}\right)^2=43+2\cdot\sqrt{29\cdot14}=43+2\cdot\sqrt{406}\)

mà 49>43 và 528>406

nên \(\left(4+\sqrt{33}\right)^2>\left(\sqrt{29}+\sqrt{14}\right)^2\)

=>\(4+\sqrt{33}>\sqrt{29}+\sqrt{14}\)

\(a,\dfrac{53}{57}=\dfrac{53.10}{57.10}=\dfrac{530}{570}=1-\dfrac{40}{570}\\ \dfrac{531}{571}=1-\dfrac{40}{571}\\ Vì:\dfrac{40}{570}>\dfrac{40}{571}\Rightarrow1-\dfrac{40}{570}< 1-\dfrac{40}{571}\\ \Leftrightarrow\dfrac{53}{57}< \dfrac{531}{571}\)

b đâu ạ

e: \(2\sqrt{26}>9\)

nên \(2\sqrt{26}+4>13\)

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

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

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

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 \(625^{10}>243^{10}\)

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

a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

a: m<n

=>2022m<2022n

b: m<n

=>-4m>-4n

a, do m<n

=> 2022m<2022n

b,do m<n

=> -4m<-4n

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