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) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

a: \(6\sqrt{3}=\sqrt{108}>\sqrt{54}=3\sqrt{6}\)

\(\Rightarrow5^{6\sqrt{3}}>5^{3\sqrt{6}}\)

b: \(\sqrt{2}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{2}+\dfrac{2}{3}}=2^{\dfrac{7}{6}}\)

\(\left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}=2^{\left(-1\right)\cdot\left(-\dfrac{4}{3}\right)}=2^{\dfrac{4}{3}}\)

mà \(\dfrac{7}{6}< \dfrac{8}{6}=\dfrac{4}{3}\).

nên \(\sqrt{2}\cdot2^{\dfrac{2}{3}}< \left(\dfrac{1}{2}\right)^{-\dfrac{4}{3}}\).

\(2\sqrt{3+\sqrt{5}}=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{2}\cdot\left(\sqrt{5}+1\right)\)

\(=\sqrt{10}+\sqrt{2}>\sqrt{10}+1\)

Vậy ....

a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)

\(\sqrt{6}< \sqrt{6,25}=2,5\); 

\(\sqrt{12}< \sqrt{12,25}=3,5\); 

\(\sqrt{20}< \sqrt{20,25}=4,5\)

=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)

Vậy P < 12

Answer:

ý a, tham khảo bài làm của @xyzquynhdi

\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)

\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

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

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

a) Ta có:

\(6\sqrt{5}=\sqrt{5\cdot36}=\sqrt{180}\)

\(5\sqrt{6}=\sqrt{6\cdot25}=\sqrt{200}\)

Mà \(\sqrt{180}< \sqrt{200}\)

Vậy: \(6\sqrt{5}< 5\sqrt{6}\)

x) Ta có: \(\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\)

Công hai vế của BĐT cho 3: 

Suy ra: \(\sqrt{8}+3< 3+3=6\)

Vậy: \(\sqrt{8}+3< 6\)

b) Ta có:

\(\sqrt{2\sqrt{3}}=\sqrt[4]{12}\)

Tương tự: \(\sqrt{3\sqrt{2}}=\sqrt[4]{18}\)

Mà \(\sqrt[4]{18}>\sqrt[4]{12}\)

Vậy.....

d) Ta có: 

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

Vậy ......

e) Ta có: 

\(\sqrt{2}-2=\frac{3\sqrt{2}-6}{3}\)

\(\sqrt{3}-3=\frac{2\sqrt{3}-6}{2}\)

Mà \(3\sqrt{2}>2\sqrt{3}\)

Vậy .....

f) ........... Đang thinking

đang dùng máy tínhmaf

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

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

Do \(2< \sqrt{2}+1\)

=> P < Q

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/ \(\sqrt[3]{54}-\sqrt[3]{16}\)

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

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

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

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