giúp mink với ik mink cần gấp lắm lun íiiiiiiiiii

a) 5, 1, 1/2, 0, -2/3, -2, -3

b) \(\sqrt{2}\), 1, 1/2, 0, -1, -4/3, \(-\sqrt{5}\), -3

a) \(1>\dfrac{1}{2}>0>-\dfrac{2}{3}>-1>-3\)

b) \(\sqrt{2}>1>\dfrac{1}{2}>0>-1>\dfrac{-4}{3}>-\sqrt{5}>-3\)

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

-2<-1.75<0<_{√5<π<22/7<5 3/6}

\(\sqrt{484}-\dfrac{1}{\sqrt{5}}< \sqrt{529}-\dfrac{1}{19}< \sqrt{576}-\dfrac{1}{\sqrt{7}}< \sqrt{625}-\dfrac{1}{\sqrt{8}}\)

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

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

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

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

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

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

0<\(\dfrac{1}{5}\) <\(\dfrac{1}{3}\) <\(\dfrac{3}{4}\) <1<\(\dfrac{4}{3}\) <2<\(\dfrac{5}{2}\) <\(\dfrac{6}{2}\)

quy đồng lên rồi sắp sếp bn

bij ngu à mà ko tự tính đi mà còn hỏi câu dễ 

mk tính đc nhưng mk ko có thời gian

`MSC:12`

`1/3 = (1xx4)/(3xx4)= 4/12`

`3/2=(3xx6)/(2xx6)=18/12`

`5/6=(5xx2)/(6xx2)=10/12`

`3/4=(3xx3)/(4xx3)=9/12`

`-> 4/12; 9/12; 10/12;18/12`

`->1/3; 3/4;5/6;3/2`

có `1/3=4/12`

`3/2=18/12`

`5/6=10/12`

`3/4=9/12`

vì `4<9<10<18`

`=>4/12<9/12<10/12<18/12`

`=>1/3<3/4<5/6<3/2`

`=>` sắp xếp: `1/3;3/4;5/6;3/2`

\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)

\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)

1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)