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 \(4\sqrt{5}>5\sqrt{3}\)

\(\sqrt{10}+\sqrt{13}< \sqrt{7}+\sqrt{17}\)

kick nhaNguyễn Đức

Bạn có biết cách giải không vậy? Giup mình với. 

Ta có \(\dfrac{3}{4}=\sqrt{\dfrac{9}{16}}< \sqrt{\dfrac{10}{17}}\Rightarrow\dfrac{3}{4}< \sqrt{\dfrac{10}{17}}\)

\(\sqrt{\dfrac{10}{17}}< \sqrt{\dfrac{9}{16}}\)

mà \(\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\)

nên \(\sqrt{\dfrac{10}{17}}< \dfrac{3}{4}\)

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

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

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

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

Giúp mình với mn

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

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

\(a\)

\(\sqrt{7}+\sqrt{15}\) 

\(=\sqrt{7+15}\)

\(=4,69\)

\(4,69< 7\)

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

\(b\)

\(\sqrt{7}+\sqrt{15}+1\)

\(=\sqrt{7+15}+1\)

\(=4,69+1\)

\(=5,69\)

\(\sqrt{45}\)

\(=6,7\)

\(5,69< 6,7\)

\(\Rightarrow\)\(\sqrt{7}+\sqrt{15}+1\)\(< \)\(\sqrt{45}\)

\(c\)

\(\frac{23-2\sqrt{19}}{3}\)

\(=\frac{22.4,53}{3}\)

\(=\frac{95,7}{3}\)

\(=31,9\)

\(\sqrt{27}\)

\(=5,19\)

\(31,9>5,19\)

\(\text{​​}\Rightarrow\text{​​}\text{​​}\)\(\frac{23-2\sqrt{19}}{3}\)\(>\sqrt{27}\)

\(d\)

\(\sqrt{3\sqrt{2}}\)

\(=\sqrt{3.1,41}\)

\(=\sqrt{4,23}\)

\(=2,05\)

\(\sqrt{2\sqrt{3}}\)

\(=\sqrt{2.1,73}\)

\(=\sqrt{3,46}\)

\(=1,86\)

\(2,05>1,86\)

\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)

\(Học \) \(Tốt !!!\)

a) Ta có : \(\sqrt{7}< \sqrt{9}=3;\sqrt{15}< \sqrt{16}=4\)

Do đó : \(\sqrt{7}+\sqrt{15}< 3+4=7\)

b) Ta có : \(\sqrt{17}>\sqrt{16}=4;\sqrt{5}>\sqrt{4}=2\)

\(\Rightarrow\sqrt{17}+\sqrt{5}+1>4+2+1=7\)

Lại có : \(\sqrt{45}< \sqrt{49}< 7\)

Do đó : \(\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)

c) Ta thấy : \(\sqrt{19}>\sqrt{16}=4\)

\(\Rightarrow2\sqrt{19}>2.4=8\)

\(\Rightarrow-2\sqrt{19}< -8\)

\(\Rightarrow23-2\sqrt{19}< 23-8=15\)

\(\Rightarrow\frac{23-2\sqrt{19}}{3}< 5\). Mặt khác : \(\sqrt{27}>\sqrt{25}=5\)

Nên : \(\frac{23-2\sqrt{19}}{3}< \sqrt{27}\)

d) Vì : \(18>12>0\Rightarrow\sqrt{18}>\sqrt{12}>0\)

\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}>0\)

\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)

\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)

\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)

\(\sqrt{2}\) + \(\sqrt{3}\)  > 2