a) \(\sqrt[3]{10}=\sqrt[15]{10^5}>\sqrt[15]{20^3=\sqrt[5]{20}}\)

b) Vì \(\frac{1}{e}<1\) và \(\sqrt{8}-3<0\) nên \(\left(\frac{1}{e}\right)^{\sqrt{8}-3}>1\)

c) Vì \(\frac{1}{8}<1\) và \(\pi>3.14\) nên \(\left(\frac{1}{8}\right)^{\pi}<\left(\frac{1}{8}\right)^{3,14}\)

d)  Vì \(\frac{1}{\pi}<1\)  và \(1,4<\sqrt{2}\)  nên \(\left(\frac{1}{\pi}\right)^{1,4}>\pi^{-\sqrt{2}}\)

dell bt

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) \(1=\sqrt{1}< \sqrt{2}\)

b) \(2=\sqrt{4}>\sqrt{3}\)

c) \(6=\sqrt{36}< \sqrt{41}\)

d) \(7=\sqrt{49}>\sqrt{47}\)

e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)

f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)

g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)

h) \(\sqrt{3}>0>-\sqrt{12}\)

i) \(5=\sqrt{25}< \sqrt{29}\)

\(\Rightarrow-5>-\sqrt{29}\)

\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)

\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)

\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)

\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)

\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)

\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)

----------------------------

\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)

\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)

\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)

\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)

Cách 1:

\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)

\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

Cách 2:

\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)

\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)

Vì $E>0$ nên $E=2$

câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :

\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)      

\(=3-\sqrt{6}+2\sqrt{6}-3\)   ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )

\(=\sqrt{6}\)

A = \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{5}+\sqrt{2}\right)}\)

A = \(\dfrac{\sqrt{3}+\sqrt{2}+3-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\) = \(\dfrac{3}{1}\) = \(3\)

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

C = \(\left(4+\sqrt{15}\right).\left(\sqrt{40-10\sqrt{15}}-\sqrt{24-6\sqrt{15}}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(\sqrt{\left(5-\sqrt{15}\right)^2}-\sqrt{\left(\sqrt{15}-3\right)^2}\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\left(\sqrt{15}-3\right)\right)\)

C = \(\left(4+\sqrt{15}\right)\left(5-\sqrt{15}-\sqrt{15}+3\right)\)

C = \(\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

C = \(32-8\sqrt{15}+8\sqrt{15}-30=2\)

D = \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

D = \(\left(\sqrt{30-10\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\left(3+\sqrt{5}\right)\)

D = \(\left(\sqrt{\left(5-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\right)\left(3+\sqrt{5}\right)\)

D = \(\left(5-\sqrt{5}-\left(\sqrt{5}-1\right)\right)\left(3+\sqrt{5}\right)\)

D = \(\left(5-\sqrt{5}-\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)

D = \(\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

D = \(18+6\sqrt{5}-6\sqrt{5}-10=8\)

E = \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{5}}\)

E = \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)

E = \(3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)