Ta có -14/21=-2/3=-2.2/3x2=-4/6

         -60/72=-5/6

Vì -5<-4

=>-5/6<-4/6

Vậy -60/12<-14/21

ta tính VT ra rồi so sánh với VP

a,Ta có:

  \(\left(\sqrt{24}+\sqrt{45}\right)^2=24+45=69\)

\(12^2=144\)

Do 69<144 nên ...

b,tương tự ý a

a ) Ta co \(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12\)

vay \(\sqrt{24}+\sqrt{45}< 12\)

b)ta co \(\sqrt{37}-\sqrt{15}>\sqrt{4}-\sqrt{0}=2-0=2\)

vay \(\sqrt{37}-\sqrt{15}>2\)

x+1/3-4=-1

=>x+1/3=-1+4

=>x+1/3=3

=>x =3-1/3

=>x =8/3

Vậy x = 8/3

(2/25-1,008):4/7:(13/4-6/5/9)*36/17

=(2/25-126/125).7/4:(13/4-59/9)*36/17

=(10/125-126/125).7/4:(117/36-236/36)*36/17

=-116/125.7/4.(-36/119).36/17

=-203/125.(-1296/2023)=263088/252875

Mình tính ko nhanh đâu

1: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}\)

\(\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)

mà 2 căn 21<4 căn 6

nên căn 3+căn 7<2+căn 6

2: \(\sqrt{7}-\sqrt{5}=\dfrac{2}{\sqrt{7}+\sqrt{5}}\)

\(\sqrt{6}-2=\dfrac{2}{\sqrt{6}+2}\)

mà \(\sqrt{7}+\sqrt{5}>\sqrt{6}+2\)

nên \(\sqrt{7}-\sqrt{5}< \sqrt{6}-2\)

3: \(\sqrt{11}-\sqrt{7}=\dfrac{4}{\sqrt{11}+\sqrt{7}}\)

\(\sqrt{7}-\sqrt{3}=\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

mà căn 11>căn 3

nên \(\sqrt{11}-\sqrt{7}< \sqrt{7}-\sqrt{3}\)

\(a,\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{3+2\sqrt{2.3}+2}-\sqrt{3-2\sqrt{2.3}+2}\)

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

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\)

\(=2\sqrt{2}\)

\(b,\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5-2\sqrt{2.5}+2}-\sqrt{5+2\sqrt{5.2}+2}\)

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

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

\(=-2\sqrt{2}\)

a) \(\sqrt{5+2\sqrt{6}}\) -\(\sqrt{5-2\sqrt{6}}\) 

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

=/\(\sqrt{3}+\sqrt{2}\)/  \(-\)/\(\sqrt{3}-\sqrt{2}\) /

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

=\(\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) 

=\(2\sqrt{2}\) 

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\) 

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

=/\(\sqrt{5}-\sqrt{2}\) / \(-\) /\(\sqrt{5}+\sqrt{2}\)/

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

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

=\(-2\sqrt{2}\)

Ta có\(\sqrt[3]{2}=1,25992105\\ \sqrt[2]{3}=1,732050808\)

=> 1,25992105<1,73205080 =>\(\sqrt[3]{2}< \sqrt[2]{3}\)

minh lan dau tien vao trang web nay nen khong biet nhieu

2003/2004 + 2004/2005 + 2005/2003

= 1 - 1/2004 + 1 - 1/2005 + 1 + 1/2003 + 1/2003

=(1+1+1)-(1/2004 - 1/2003 + 1/2005 - 1/2003)

= 3 - (1/2004 - 1/2003 + 1/2005 - 1/2003)

Vì 1/2004 < 1/2003 ; 1/2005 < 1/2003

=>1/2004 - 1/2003 + 1/2005 - 1/2003 < 0

=> 3 - (...) > 3

Vậy. ...

K mình nha