a: m>n

=>2m>2n

=>2m-2>2n-2

b: m>n

=>-3m<-3n

=>-3m+1<-3n+1

c: m>n

=>2m>2n

=>2m+3>2n+3

mà 2n+3>2n+1

nên 2m+3>2n+1

d: m>n

=>-5m<-5n

=>-5m+3<-5n+3

mà -5n+3<-5n+7

nên -5m+3<-5n+7

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

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

\(b,\dfrac{25}{26}=\dfrac{25.1010}{26.1010}=\dfrac{25250}{26260}=1-\dfrac{1010}{26260}\\ \dfrac{25251}{26261}=1-\dfrac{1010}{26261}\\ Vì:\dfrac{1010}{26260}>\dfrac{1010}{26261}\Rightarrow1-\dfrac{1010}{26260}< \dfrac{1010}{26261}\\ \Leftrightarrow\dfrac{25}{26}< \dfrac{25251}{26261}\)

b: \(\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{7}+2-\sqrt{7}+\sqrt{3}=2+\sqrt{3}\)

a,333^444 < 444^333

B,5^340 > 7^255

\(1.\)

a, \(27^{265}\)và \(81^{199}\)

\(27^{265}=\left(3^3\right)^{265}=3^{795}\)

\(81^{199}=\left(3^4\right)^{199}=3^{796}\)

\(\Rightarrow3^{795}< 3^{796}hay27^{265}< 81^{199}\)

b, \(1024^{15}=\left(2^{10}\right)^{15}=2^{150}\)

\(128^{21}=\left(2^7\right)^{21}=2^{147}\)

\(2^{150}>2^{147}.hay.1024^{15}>128^{21}\)

A<B

Tham khảo: https://hoc24.vn/cau-hoi/so-sanh-afrac720131720141va-bfrac720141720151.235038064198

a)2^6=8^2

b)5^3<3^5

a) \(2^6\) và \(8^2\)

   \(2^6=\left(2^2\right)^3\)

  \(8^2=\left(2^3\right)^2\)\(=2^6\)

 \(\Rightarrow\) \(2^6=8^2\)