b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)

\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có  \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)

\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)

So sánh gì thế em, em nhập đủ đề vào hi

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

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

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

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

Bài 2 : 

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

b. \(\sqrt{37}-\sqrt{15}>\sqrt{36}-\sqrt{16}=6-4=2=>\sqrt{37}-\sqrt{15}>2\)

c, \(\sqrt{15}.\sqrt{17}>\sqrt{15}.\sqrt{16}>\sqrt{16}=>\sqrt{15}.\sqrt{17}>\sqrt{16}\)

a    \(\left(\sqrt{5\sqrt{7}}\right)^4=\left(\left(\sqrt{5\sqrt{7}}\right)^2\right)^2=\left(5\sqrt{7}\right)^2=25\cdot7=175\)

\(=\left(\sqrt{7\sqrt{5}}\right)^4=\left(\left(\sqrt{7\sqrt{5}}\right)^2\right)^2=\left(7\sqrt{5}\right)^2=49\cdot5=240\)

vì 175<240\(\Rightarrow\left(\sqrt{5\sqrt{7}}\right)^4< \left(\sqrt{7\sqrt{5}}\right)^4\Rightarrow\sqrt{5\sqrt{7}}< \sqrt{7\sqrt{5}}\)

b     \(6=\sqrt{36}\)

\(\sqrt{31}< \sqrt{36};\sqrt{19}>\sqrt{17}\Rightarrow\sqrt{31}-\sqrt{19}< \sqrt{36}-\sqrt{17}=6-\sqrt{17}\)

c      \(\left(\sqrt{10}+\sqrt{17}\right)^2=10+2\sqrt{10\cdot17}+17=27+2\sqrt{170}\)

\(\left(\sqrt{61}\right)^2=61=27+34=27+2\cdot17=27+2\sqrt{289}\)

vì \(2\sqrt{170}< 2\sqrt{289}\Rightarrow27+2\sqrt{170}< 27+2\sqrt{289}\Rightarrow\left(\sqrt{10}+\sqrt{17}\right)^2< \left(\sqrt{61}\right)^2\)

\(\Rightarrow\sqrt{10}+\sqrt{17}< \sqrt{61}\)

So sánh 

\(a,\sqrt{91}>9\)

\(b,3>\sqrt{5-1}\)

\(c,5\sqrt{17}>20\)

\(d,\sqrt{7}+\sqrt{15}< 7\)

a) Ta có 290>289

<=>  \(\sqrt{290}\)   >       \(\sqrt{289}\)

<=>  \(\sqrt{290}\)   >        17

Vậy ..........

\(a,290>289\)

\(\Rightarrow\sqrt{290}>\sqrt{289}\)

\(\Rightarrow\sqrt{290}>17\)

\(b,\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< 3+4\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)

Diệp ơi hỏi bài tập toán nhà thầy long hở kinh nhờ