\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

\(A=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)

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

mà \(\sqrt{12}+\sqrt{11}< \sqrt{14}+\sqrt{13}\)

nên A>B

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Mình chọn nhầm lớp 8 chứ thật ra câu hỏi ở bên lớp 9 

a) Ta có \(5=\sqrt{25}\)

Vì \(\sqrt{25}>\sqrt{11}\) nên \(5>\sqrt{11}\)

b) Ta có \(4=\sqrt{16}\)

Vì \(\sqrt{13}< \sqrt{16}\) nên \(\sqrt{13}< 4\)

c) Ta có \(-7=-\sqrt{49}\)

Vì \(-\sqrt{49}< -\sqrt{43}\) nên \(-7< -\sqrt{43}\)

d) Ta có \(-5=-\sqrt{25}\)

Vì \(-\sqrt{21}>-\sqrt{25}\) nên \(-\sqrt{21}>-5\)

a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)

=> A²=8+2\(\sqrt{3}\)

B=\(\sqrt{3}+1\)=> B²=10+2\(\sqrt{3}\)

=>A>B

a: \(\sqrt{a^2}=\left|a\right|\)

\(\sqrt[3]{a^3}=a\)

b: \(\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\)

a] < b] < c] >

ko chi tiết lắm

\(\frac{13-2\sqrt{3}}{6}< 2\)

Ta có :\(-2\sqrt{3}>-2\sqrt{4}=-4\)   =>\(-2\sqrt{3}>-4\)

\(\frac{13-2\sqrt{3}}{6}>\frac{13-4}{6}=\frac{3}{2}=\sqrt{\frac{9}{4}}>\sqrt{\frac{8}{4}}=\sqrt{2}\)

<=>\(\frac{13-2\sqrt{3}}{6}>\sqrt{2}\)