a,\(\sqrt{105}vs\sqrt{101}\) b,\(\sqrt{101}vs\sqrt{97}\)

Có:105>101 Có:101>97

=>\(\sqrt{105}>\sqrt{101}\) =>\(\sqrt{101}>\sqrt{97}\)

a,\(\sqrt{105}vs\sqrt{101}\)

Có:105>101

=>\(\sqrt{105}>\sqrt{101}\)

b,\(\sqrt{101}vs\sqrt{97}\)

Có:101>97

=>\(\sqrt{101}>\sqrt{97}\)

\(A=\sqrt{15}-\sqrt{14}=\dfrac{1}{\sqrt{15}+\sqrt{14}}\)

\(B=\sqrt{14}-\sqrt{13}=\dfrac{1}{\sqrt{14}+\sqrt{13}}\)

hiển nhiên

\(\sqrt{15}+\sqrt{14}>\sqrt{14}+\sqrt{13}\)

\(=>A< B\)

Với n\(\in\)N thì \(\frac{1}{\sqrt{n+4}+\sqrt{n}}=\frac{\sqrt{n+4}-\sqrt{n}}{n+4-n}\)\(=\frac{\sqrt{n+4}-\sqrt{n}}{4}\)

\(\Leftrightarrow\frac{4}{\sqrt{n+4}+\sqrt{n}}=\sqrt{n+4}-\sqrt{n}\) (1)

Áp dụng bất đẳng thức (1) ta được:

\(\sqrt{105}-\sqrt{101}=\frac{4}{\sqrt{105}+\sqrt{101}}\)

\(\sqrt{101}-\sqrt{97}=\frac{4}{\sqrt{101}+\sqrt{97}}\)

Ta thấy: \(\sqrt{105}+\sqrt{101}>\sqrt{101}+\sqrt{97}\)

\(\Leftrightarrow\frac{4}{\sqrt{105}+\sqrt{101}}< \frac{4}{\sqrt{101}+\sqrt{97}}\) hay \(\sqrt{105}-\sqrt{101}< \sqrt{101}-\sqrt{97}\)

Vậy bất đẳng thức được chứng minh.

a) Ta thấy:
\(\left(3+\sqrt{5}\right)^2=\left(\sqrt{9}+\sqrt{5}\right)^2=9+5+2\sqrt{45}=14+2\sqrt{45}\)
\(\left(2\sqrt{2}+\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2=8+6+2\sqrt{48}=14+2\sqrt{48}\)
Vì \(45< 48\)
\(\Rightarrow\sqrt{45}< \sqrt{48}\)
\(\Rightarrow2\sqrt{45}< 2\sqrt{48}\)
\(\Rightarrow14+2\sqrt{45}< 14+2\sqrt{48}\)
\(\Rightarrow\left(3+\sqrt{5}\right)^2< \left(2\sqrt{2}+\sqrt{6}\right)^2\)
Do \(3+\sqrt{5}>0;2\sqrt{2}+\sqrt{6}>0\)
\(\Rightarrow3+\sqrt{5}< 2\sqrt{2}+6\)

b) Ta thấy:
Vì \(26>3\)
\(\Rightarrow\sqrt{26}>\sqrt{3}\)
\(\Rightarrow\sqrt{26}+1>\sqrt{3}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{27}+\sqrt{3}\)
Mà \(\sqrt{27}+\sqrt{3}=3\sqrt{3}+\sqrt{3}=4\sqrt{3}=\sqrt{48}\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{48}\)

\(P=\sqrt{101-2\sqrt{101}+1}+\sqrt{101+2\sqrt{101}+1+1}\)

    \(=\sqrt{\left(\sqrt{101}-1\right)^2}+\sqrt{\left(\sqrt{101}+1\right)^2+1}>\sqrt{101}-1+\sqrt{101}+1=2\sqrt{101}>2.\sqrt{100}=2.10=20\)

=> P > 20

\(\sqrt{99}+\sqrt{101}=9,94........+10,04.......\)

Mà 9,94 + 10,04 = 19,98 < 20

Vậy \(\sqrt{99}+\sqrt{101}< 20\)

Xét : (\(\sqrt{99}+\sqrt{101}\))^2 = 99+101 + 2\(\sqrt{99.101}\)<= 200 + 99+101 ( bđt cosi ) = 400

=> \(\sqrt{99}+\sqrt{101}\)< 20

k mk nha

Ý anh là so sánh đúng ko ạ?

15) Bình phương hai vế,ta cần so sánh: \(\left(\frac{5}{4}\sqrt{2}\right)^2\text{ và }\left(\frac{2}{3}\sqrt{7}\right)^2\Leftrightarrow\frac{25}{8}\text{ và }\frac{28}{9}\)

Dễ thấy \(\frac{25}{8}>\frac{28}{9}\Rightarrow\frac{5}{4}\sqrt{2}>\frac{2}{3}\sqrt{7}\)

16) \(\sqrt{15}-\sqrt{14}=\frac{1}{\sqrt{15}+\sqrt{14}}< \frac{1}{\sqrt{14}+\sqrt{13}}=\sqrt{14}-\sqrt{13}\)

Xíu em làm tiếp,tắm đã

17/ Tương tự câu 16,18

18) \(\sqrt{9}-\sqrt{7}=\frac{2}{\sqrt{9}+\sqrt{7}};\sqrt{7}-\sqrt{5}=\frac{2}{\sqrt{7}+\sqrt{5}}\)

Dễ thấy \(\sqrt{9}+\sqrt{7}>\sqrt{7}+\sqrt{5}\Rightarrow\sqrt{9}-\sqrt{7}< \sqrt{7}-\sqrt{5}\)

13)Ta có: \(2\sqrt{6}=\sqrt{4.6}=\sqrt{24}>\sqrt{23}\Rightarrow-2\sqrt{6}< -\sqrt{23}\)

14)\(\sqrt{111}-7< \sqrt{121}-7=11-7=4\)

:v Thứ tự ngộ nhỉ?

Ta có : \(\frac{1}{\sqrt{k}+\sqrt{k+1}}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

Áp dụng : A = 2\(\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)=  \(2\left(\sqrt{101}-1\right)\) \(\ge\) \(2\left(\sqrt{100}-1\right)=2\left(10-1\right)=2\times9=18\) 

B = \(\frac{181}{20}=9,05\) < 18 nên suy ra : A>B