\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)

\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)

\(\Rightarrow A< B\)

Ta có:

 \(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)

\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)

Lại có :

\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)

Mà B > 0

\(\Rightarrow B>7\) (2)

Từ (1),(2) suy ra A<B

ta xét hiệu A - B= \(\left(\sqrt{10}+\sqrt{13}\right)-\left(\sqrt{11}+\sqrt{12}\right)\) = \(\left(\sqrt{13}-\sqrt{12}\right)-\left(\sqrt{11}-\sqrt{10}\right)\)

\(\le\sqrt{13-12}-\sqrt{11-10}=1-1=0\)

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 \(-\sqrt{21}>-5\)

tính

\(\frac{a-\sqrt{ab}}{b-\sqrt{ab}}+\frac{b-\sqrt{ab}}{a+\sqrt{ab}}=\frac{a-ab+b-ab}{ab+b\sqrt{ab}-a\sqrt{ab}-ab}=\frac{a+b}{\sqrt{ab}\left(b-a\right)}\)

còn lại mk chịu

bạn ghi rõ hơn nữa được không chứ mình chưa hiểu lắm

a)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11

a) Ta có : \(\left(\sqrt{11}+\sqrt{13}\right)^2=11+2\sqrt{11.13}+13=24+2\sqrt{143}\)

\(\left(2.\sqrt{12}\right)^2=4.12=24+2.\sqrt{144}\)

mà \(\sqrt{144}>\sqrt{143}\Rightarrow24+2\sqrt{144}>24+2\sqrt{143}\Rightarrow\left(2.\sqrt{12}\right)^2>\left(\sqrt{11}+\sqrt{13}\right)^2\)

\(2.\sqrt{12}>\sqrt{11}+\sqrt{13}\)

b) Ta có : \(\left(\sqrt{69}-\sqrt{68}\right)-\left(\sqrt{68}-\sqrt{69}\right)\)

        \(\Leftrightarrow\sqrt{69}+\sqrt{67}-2\sqrt{68}\)

Từ kq câu a \(\Rightarrow\sqrt{69}+\sqrt{67}< 2\sqrt{68}\)

\(\Rightarrow\sqrt{69}+\sqrt{67}-2\sqrt{68}< 0\)

\(\Rightarrow\left(\sqrt{69}-\sqrt{68}\right)-\left(\sqrt{68}-\sqrt{67}\right)< 0\)

\(\Rightarrow\sqrt{69}-\sqrt{68}< \sqrt{68}-\sqrt{67}\)

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

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

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)