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; \( - \sqrt {2,3} ; - \sqrt {2\frac{1}{3}} \)

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

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

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

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

mà 96>90>80>75

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

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

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

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

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

mà 96>90>80>75

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

\(\sqrt{16}=4;\dfrac{2}{3}=0,\left(6\right);\Omega=3,14;-\sqrt{5}\simeq-2,24\)

\(-5,6< -2,23< 0\)

=>\(-5,6< -\sqrt{5}< 0\)(1)

\(0< \dfrac{2}{3}< 3,14< 4\)

=>\(0< \dfrac{2}{3}< \Omega< \sqrt{16}\)(2)

Từ (1) và (2) suy ra \(-5,6< -\sqrt{5}< 0< \dfrac{2}{3}< \Omega< \sqrt{16}\)

Ta có:

\(-\frac{2}{3} = -0,\left( 6 \right);\,\,\,\,\,4,1;\,\,\, - \sqrt 2  =  - 1,414...;\,\,\,\,3,2;\\\pi  = 3,141...;\,\,\,\, - \frac{3}{4} =  - 0,75;\,\,\,\,\frac{7}{3} = 2,\left( 3 \right)\).

Do \( - 1,414... <  - 0,75 < -0,\left( 6 \right) < 2,\left( 3 \right) < 3,141... < 3,2 < 4,1\)

Nên \( - \sqrt 2  <  - \frac{3}{4} < -\frac{2}{3} < \frac{7}{3} < \pi  < 3,2 < 4,1.\)

Ta có: 

\(-\dfrac{2}{3}\approx-0,67;-\sqrt{2}\approx-1,41;-\dfrac{3}{4}=-0,75;\dfrac{7}{3}\approx2,33;\pi\approx3,14\)

Từ đó, ta có thứ tự sắp xếp: 

\(-\sqrt{2};-\dfrac{3}{4};-\dfrac{2}{3};1;2;\dfrac{7}{3};3;\pi;4\)

\(\left| { - 3,2} \right| = 3,2;\,\,\,\,\,\left| {2,13} \right| = 2,13;\,\,\,\left| {\, - \sqrt 2 } \right| = \sqrt 2  = 1,41..;\,\,\,\,\left| { - \frac{3}{7}} \right| = \frac{3}{7} = 0,42...\)

Do \(0,42 < 1,41... < 2,13 < 3,2\) nên:

\(\left| { - \frac{3}{7}} \right| < \left| { - \sqrt 2 } \right| < \left| {2,13} \right| < \left| { - 3,2} \right|\).

\(-2< -1,75< 0< \sqrt{5}< \pi< \dfrac{22}{7}< 5\dfrac{3}{6}.\)

-2<-1.75<0<_{√5<π<22/7<5 3/6}