a) Ta sắp xếp theo thứ tự tăng dần như sau:

\(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)

b) Ta sắp xếp theo thứ tự tăng dần như sau:

\(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)

Hoàng Phong làm bừa

a/

\(2\sqrt{6}=\sqrt{24}< \sqrt{29}< 4\sqrt{2}=\sqrt{32}< 3\sqrt{5}=\sqrt{45}.\)

b/

\(\sqrt{38}< 2\sqrt{14}=\sqrt{56}< 3\sqrt{7}=\sqrt{63}< 6\sqrt{2}=\sqrt{72}\)

a. \(3\sqrt{5}=\sqrt{45}\) ; \(2\sqrt{6}=\sqrt{24}\) ; \(4\sqrt{2}=\sqrt{32}\)

Vì 24 < 29 < 32 < 45 nên \(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)

Hay \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)

b. \(6\sqrt{2}=\sqrt{72}\) ; \(3\sqrt{7}=\sqrt{63}\) ; \(2\sqrt{14}=\sqrt{56}\)

Vì 38 < 56 < 63 < 72 nên \(\sqrt{38}< \sqrt{56}< \sqrt{63}< \sqrt{72}\)

Hay \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)

A trước b sau

a) Ta có:

\(6 = \sqrt {36} ; - 1,7 =  - \sqrt {2,89} \)

Vì 0 < 2,89 < 3 nên 0> \( - \sqrt {2,89}  >  - \sqrt 3 \) hay 0 > -1,7 > \( - \sqrt 3 \)

Vì 0 < 35 < 36 < 47  nên \(0 < \sqrt {35}  < \sqrt {36}  < \sqrt {47} \) hay 0 < \(\sqrt {35}  < 6 < \sqrt {47} \)

Vậy các số theo thứ tự tăng dần là: \( - \sqrt 3 ; - 1,7;0;\sqrt {35} ;6;\sqrt {47} \)

b) Ta có:

\(\sqrt {5\frac{1}{6}}  = \sqrt {5,1(6)} ; - \sqrt {2\frac{1}{3}}  =  - \sqrt {2,(3)} \); -1,5 = \( - \sqrt {2,25} \)

Vì 0 < 2,25 < 2,3 < 2,(3) nên 0> \( - \sqrt {2,25}  >  - \sqrt {2,3}  >  - \sqrt {2,(3)} \) hay 0 > -1,5 > \( - \sqrt {2,3}  >  - \sqrt {2\frac{1}{3}} \)

Vì 5,3 > 5,1(6) > 0 nên \(\sqrt {5,3}  > \sqrt {5,1(6)} \)> 0 hay \(\sqrt {5,3}  > \sqrt {5\frac{1}{6}}  > 0\)

Vậy các số theo thứ tự giảm dần là: \(\sqrt {5,3} ;\sqrt {5\frac{1}{6}} ;0\); -1,5; \( - \sqrt {2,3} ; - \sqrt {2\frac{1}{3}} \)

a) 2 \sqrt{6}, \sqrt{29}, 4 \sqrt{2}, 3 \sqrt{5} ;26​,29​,42​,35​;

b) \sqrt{38}, 2 \sqrt{14}, 3 \sqrt{7}, 6 \sqrt{2}38​,214​,37​,62

a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)

b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)

a, \(2\sqrt{6}\),\(\sqrt{29}\),\(4\sqrt{2}\),\(3\sqrt{5}\)          

b,\(\sqrt{38}\),\(2\sqrt{14}\),\(3\sqrt{7}\),\(6\sqrt{2}\)

a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)

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

=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)

=-7

b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)

So sánh:

1) \(2\sqrt{27}\) và \(\sqrt{147}\)

+ \(2\sqrt{27}\) = \(6\sqrt{3}\)

+ \(\sqrt{147}\) = \(7\sqrt{3}\)

⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)

Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)

2) \(2\sqrt{15}\) và \(\sqrt{59}\)

+ \(2\sqrt{15}\) = \(\sqrt{60}\)

⇒ \(\sqrt{60}\) > \(\sqrt{59}\)

Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)

3) \(2\sqrt{2}-1\) và 2

\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)

So sánh: 3 và \(2\sqrt{2}\)

+ 3 = \(\sqrt{9}\)

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

⇒ \(\sqrt{8}\) < \(\sqrt{9}\)

⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1

⇒ \(2\sqrt{2}\) - 1 < 3 - 1

Vậy: \(2\sqrt{2}-1< 2\)

4) \(\frac{\sqrt{3}}{2}\) và 1

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

⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)

Vậy: \(\frac{\sqrt{3}}{2}\) < 1

5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)

+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)

⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)

Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)

Ta có :

\(2\sqrt{3}=\sqrt{12}\)

\(5\sqrt{2}=\sqrt{50}\)

\(3\sqrt{2}=\sqrt{18}\)

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

\(\Rightarrow\) \(\sqrt{12}< \sqrt{18}< \sqrt{20}< \sqrt{50}\)

Sắp xếp theo tt tăng dần : \(2\sqrt{3}< 3\sqrt{2}< 2\sqrt{5}< 5\sqrt{2}\)

\(\left(3\sqrt{10}\right)^2=90\)

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

\(\left(4\sqrt{5}\right)^2=80\)

\(\left(12\sqrt{\dfrac{2}{3}}\right)^2=96\)

mà 96>90>80>75

nên \(12\sqrt{\dfrac{2}{3}}>3\sqrt{10}>4\sqrt{5}>5\sqrt{3}\)