theo ket qua cho thay:9.4594<10

Ta có :

\(\sqrt{3}< \sqrt{4}=2\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{24}< \sqrt{25}=5\)

\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)

Vậy ...

\(\sqrt{3}+\sqrt{8}+\sqrt{24}< \sqrt{4}+\sqrt{9}+\sqrt{25}\)

\(=2+3+5=10\)

Vậy: \(\sqrt{3}+\sqrt{8}+\sqrt{24}< 10\)

\(\sqrt{27}-\sqrt{12}-\sqrt{2016}>\sqrt{25}-\sqrt{16}-\sqrt{2025}\)

\(=5-4-45=-44\)

Vậy \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

Có : \(\sqrt{12}< \sqrt{16}=4\)

         \(\sqrt{2016}< \sqrt{2025}\)         => \(\sqrt{12}+\sqrt{2016}< 4+45\)

                                                                 => \(-\sqrt{12}-\sqrt{2016}>-49\)(1)

Lại có : \(\sqrt{27}>\sqrt{25}=5\)(2)

Từ (1),(2) có : \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>5-49\)or \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

\(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)

Vậy \(\sqrt{35}+\sqrt{99}< 16\)

\(x^2=3+5+2\sqrt{15}=8+\sqrt{60}\)

\(y^2=2+6+2\sqrt{12}=8+\sqrt{48}\)

Mà \(60>48\Rightarrow\sqrt{60}>\sqrt{48}\Rightarrow8+\sqrt{10}>8+\sqrt{48}\)

\(\Rightarrow x^2>y^2\Rightarrow x>y\) (do x;y đều dương)

Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

bình phương 2 vế ta có:

vế 1 bằng 50+2=52

vế 2 bằng 50+ 10+ 2 = 62

vậy (1) < (2)

vẫn k hiu hả

\(\left(\sqrt{50+2}\right)^2=52\)

\(\left(\sqrt{50}+\sqrt{2}\right)^2=62\)

\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

2017>2015

=>căn 2017>căn 2015

=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)

=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)

\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)

\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)

\(2\sqrt{3}=\sqrt{12}< \sqrt{18}=3\sqrt{2}\)

=>\(2^{2\sqrt{3}}< 2^{3\sqrt{2}}\)