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\(\left| { - 3,2} \right| = 3,2;\,\,\,\,\,\left| {2,13} \right| = 2,13;\,\,\,\left| {\, - \sqrt 2 } \right| = \sqrt 2 = 1,41..;\,\,\,\,\left| { - \frac{3}{7}} \right| = \frac{3}{7} = 0,42...\)
Do \(0,42 < 1,41... < 2,13 < 3,2\) nên:
\(\left| { - \frac{3}{7}} \right| < \left| { - \sqrt 2 } \right| < \left| {2,13} \right| < \left| { - 3,2} \right|\).
\(-2< -1,75< 0< \sqrt{5}< \pi< \dfrac{22}{7}< 5\dfrac{3}{6}.\)
a: -10/8<-1
-19/19=-1
-1<-2/10<0
0<5/12<1<17/15
=>17/15>1>5/12>-2/10>-19/19>-10/8
b: -1/3=-4/12; -5/12=-5/12; -3/4=-9/12; -1/4=-3/12; -7/12=-7/12
=>-3/4<-7/12<-5/12<-1/3<-1/4
a: \(-3< -2.15< -\sqrt{3}< 0< \dfrac{13}{7}< \sqrt{8}< \dfrac{33}{12}\)
b: \(0< \sqrt{3}< \dfrac{13}{7}< 2.15< \dfrac{33}{12}< \sqrt{8}< 3\)
a)-3<-2<-\(\sqrt[]{3}\)<0<\(\dfrac{13}{7}\)<\(\dfrac{33}{12}\)<\(\sqrt{8}\)<15
b)|0|<|-\(\sqrt{3}\)|\(\dfrac{13}{7}\)|<|-2|<|\(\dfrac{33}{12}\)|<\(\sqrt{8}\)<|-3|<15
\(\sqrt{16}=4;\dfrac{2}{3}=0,\left(6\right);\Omega=3,14;-\sqrt{5}\simeq-2,24\)
\(-5,6< -2,23< 0\)
=>\(-5,6< -\sqrt{5}< 0\)(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:
\(-\sqrt{2};-\dfrac{3}{4};-\dfrac{2}{3};1;2;\dfrac{7}{3};3;\pi;4\)