\(5^{30}=\left(5^3\right)^{10}=125^{10}\)

Vì 125¹⁰>124¹⁰=>5³⁰>124¹⁰

a,

\(5^{30}\)và  \(124^{10}\)

\(5^{30}=\left(5^3\right)^{10}=125^{10}\)

Vì \(125^{10}>124^{10}\)nên \(5^{30}>124^{10}\)

b, \(31^{11}\)và  \(17^{14}\)

\(31^{11}< 32^{11}=\left(2^5\right)^{11}=2^{55}\)

\(17^{14}< 16^{14}=\left(2^4\right)^{14}=2 ^{56}\)

Vì \(2^{55}< 2^{56}\)nên \(31^{11}< 17^{14}\)

3¹¹ = 177147, 17¹⁴ = 1.68377826...

3^11 < 17^14

124^10 > 5^21

Hết ................

a)ta thấy 3<17;11<14 =>3¹¹<17¹⁴

`@` `\text {Ans}`

`\downarrow`

`\sqrt {2} + \sqrt {11}` và `\sqrt {3} + 5`

Ta có: `5^2 = 25`

`=> \sqrt {25} = 5`

`=> \sqrt {3} + 5 = \sqrt {3} + \sqrt {25}`

Vì: \(\left\{{}\begin{matrix}\sqrt{3}>\sqrt{2}\\\sqrt{25}>\sqrt{11}\end{matrix}\right.\)

`=>`\(\sqrt{3}+\sqrt{25}>\sqrt{2}+\sqrt{11}\)

`=> \sqrt {3} + 5 > \sqrt {2} + \sqrt {11}.`

`# \text {NgMH}`

(căn 2+căn 11)^2=13+2*căn 22

(căn 3+5)^2=28+2*căn 45

mà 13<28; căn 22<căn 45

nên căn 2+căn 11<căn 3+5

a)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11

a: \(1< \sqrt{2}\)

nên \(2< \sqrt{2}+1\)

b: \(2\sqrt{31}=\sqrt{124}\)

\(10=\sqrt{100}\)

mà 124>100

nên \(2\sqrt{31}>10\)

c: \(-3\sqrt{11}=-\sqrt{99}\)

\(-\sqrt{12}=-\sqrt{12}\)

mà 99>12

nên \(-3\sqrt{11}< -\sqrt{12}\)

\(-\frac{11}{25}>-\frac{25}{25}\) ; \(-\frac{25}{25}>-\frac{25}{124}\)

Suy ra \(-\frac{11}{25}>-\frac{25}{124}\)

11/-25 =-1364/3100

-25/124=-625/3100

do -1364/3100<-625/3100 suy ra 11/-25<-25/124

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

\(A=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

\(B=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

mà \(\sqrt{12}+\sqrt{11}< \sqrt{14}+\sqrt{13}\)

nên A>B

Mình chọn nhầm lớp 8 chứ thật ra câu hỏi ở bên lớp 9 

a) Ta có \(5=\sqrt{25}\)

Vì \(\sqrt{25}>\sqrt{11}\) nên \(5>\sqrt{11}\)

b) Ta có \(4=\sqrt{16}\)

Vì \(\sqrt{13}< \sqrt{16}\) nên \(\sqrt{13}< 4\)

c) Ta có \(-7=-\sqrt{49}\)

Vì \(-\sqrt{49}< -\sqrt{43}\) nên \(-7< -\sqrt{43}\)

d) Ta có \(-5=-\sqrt{25}\)

Vì \(-\sqrt{21}>-\sqrt{25}\) nên \(-\sqrt{21}>-5\)

\(5=2+3\)

\(=2+\sqrt{9}< 2+\sqrt{11}\)

Giả sử \(5< 2+\sqrt[]{11}\)

\(\Leftrightarrow3< \sqrt[]{11}\)

\(\Leftrightarrow9< 11\left(luôn.đúng\right)\)

Vậy \(5< 2+\sqrt[]{11}\)