\(a,ƯCLN\left(540,168\right)=12\\ \RightarrowƯC\left(540,168\right)=Ư\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\\ b,ƯCLN\left(735,350\right)=35\\ \RightarrowƯC\left(735,350\right)=Ư\left(35\right)=\left\{-35;-7;-5;-1;1;5;7;35\right\}\)

\(a.540=2^2.3^3.5\)

\(168=2^3.3.7\)

\(ƯCLN\left(540;168\right)=2^2.3=12\)

\(ƯC\left(540;168\right)=\left\{12;24;36;...\right\}\)

\(b.735=3.5.7^2\)

\(350=2.5^2.7\)

\(ƯCLN\left(735;350\right)=5.7=35\)

\(ƯC\left(735;350\right)=\left\{35;70;105;...\right\}\)

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

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

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

a) \(5^{48}=\left(5^4\right)^{12}=625^{12}\)

\(2^{108}=\left(2^9\right)^{12}=512^{12}\)

Do \(625>512\Rightarrow625^{12}>512^{12}\) \(\Rightarrow5^{48}>2^{108}\)  (1)

Lại có: \(108>105\Rightarrow2^{108}>2^{105}\)   (2)

Từ (1) và (2) \(\Rightarrow5^{48}>2^{105}\)

b) \(2^{50}=\left(2^5\right)^{10}=32^{10}\)

Do \(33>32\Rightarrow33^{10}>32^{10}\)

Vậy \(33^{10}>2^{50}\)

c) Do \(513>512\Rightarrow513^{100}>512^{100}\)   (1)

\(512^{100}=\left(2^9\right)^{100}=2^{900}\) \(=2^{10.90}=\left(2^{10}\right)^{90}=1024^{90}\) (2)

Do \(1024>1023\Rightarrow1024^{90}>1023^{90}\) (3)

Từ (1), (2) và (3) \(\Rightarrow513^{100}>1023^{90}\)

a) 0,(26)<0,261

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

a: 0,(26)<0,261

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

a: m<n

=>2022m<2022n

b: m<n

=>-4m>-4n

a, do m<n

=> 2022m<2022n

b,do m<n

=> -4m<-4n

a) \(2=\sqrt{4}>\sqrt{3}\)

b) \(6=\sqrt{36}< \sqrt{41}\)

c) \(7=\sqrt{49}>\sqrt{47}\)

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

a: \(2\sqrt{2}< 3\)

nên \(6+2\sqrt{2}< 9\)

a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)

b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)

a: \(2\sqrt{5}=\sqrt{20}\)

\(5=\sqrt{25}\)

mà 20<25

nên \(2\sqrt{5}< 5\)

b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)

\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)

mà 16<9

nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)