\(\sqrt{3\sqrt{3}}=\sqrt{3.\sqrt{\dfrac{432}{144}}}< \sqrt{3\sqrt{\dfrac{625}{144}}}=\sqrt{3.\dfrac{25}{12}}=\dfrac{5}{2}\)

\(\sqrt{3}+1=\sqrt{\dfrac{12}{4}}+1>\sqrt{\dfrac{9}{4}}+1=\dfrac{3}{2}+1=\dfrac{5}{2}\)

\(\Rightarrow\sqrt{3}+1>\sqrt{3\sqrt{3}}\)

a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)

\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)

Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)

b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)

\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)

Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)

\(P=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

\(Q=\dfrac{1}{\sqrt{2}-1}=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)

Do \(2< \sqrt{2}+1\)

=> P < Q

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

2/ 

a) Ta có:

\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)

\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)

Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)

b) Ta có:

\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)

\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)

Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)

3/

a)ĐKXĐ: \(x\ne1;x\ge0\)

b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)

\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)

\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)

\(A=1^2-\left(\sqrt{x}\right)^2\)

\(A=1-x\)

1/ \(\sqrt[3]{54}-\sqrt[3]{16}\)

\(=\sqrt[3]{3^3\cdot2}-\sqrt[3]{2^3\cdot2}\)

\(=3\sqrt[2]{3}-2\sqrt[3]{2}\)

\(=\left(3-2\right)\sqrt[3]{2}\)

\(=\sqrt[3]{2}\)

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

\(1.-3< -5+\sqrt{5}\)

\(2.-4>-2\sqrt{5}\)

\(3.-3\sqrt{5}< -6\)

2) \(4=\sqrt{16}\)

\(2\sqrt{5}=\sqrt{20}\)

mà 16<20

nên \(-4>-2\sqrt{5}\)

3) \(3\sqrt{5}=\sqrt{45}\)

\(6=\sqrt{36}\)

mà 45>36

nên \(-3\sqrt{5}< -6\)

1)Ta có \(-3=-\sqrt{9}>-5+\sqrt{5}\)

2)Ta có \(-2\sqrt{5}=(-\sqrt{20})<-4=(-\sqrt{16})\)

3)Ta có \(-3\sqrt{5}=(-\sqrt{45})<-6=-\sqrt{36}\)

a) Ta có:

\(2=1+1=1+\sqrt{1}\)

Mà: \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\)

\(\Rightarrow1+\sqrt{1}< \sqrt{2}+1\)

\(\Rightarrow2< \sqrt{2}+1\)

b) Ta có:

\(1=2-1=\sqrt{4}-1\)

Mà: \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\)

\(\Rightarrow\sqrt{4}-1>\sqrt{3}-1\)

\(\Rightarrow1>\sqrt{3}-1\)

c) Ta có:

\(10=2\cdot5=2\sqrt{25}\)

Mà: \(25< 31\Rightarrow\sqrt{25}< \sqrt{31}\)

\(\Rightarrow2\sqrt{25}< 2\sqrt{31}\)

\(\Rightarrow10< 2\sqrt{31}\)

d) Ta có:

\(-12=-3\cdot4=-3\sqrt{16}\)

Mà: \(16>11\Rightarrow\sqrt{16}>\sqrt{11}\)

\(\Rightarrow-3\sqrt{16}< -3\sqrt{11}\)

\(\Rightarrow-12< -3\sqrt{11}\)