Tính: \(\sqrt{106-24\sqrt{ }10}+\dfrac{30}{\sqrt{10}}-\dfrac{12}{\sqrt{10}+2}\)
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18 tháng 10 2021
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
3 tháng 7 2017
\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)
Ta có : \(6< 6.25\Rightarrow\sqrt{6}< \sqrt{6.25}\Rightarrow\sqrt{6}< 2.5\)
\(12< 12.25\Rightarrow\sqrt{12}< \sqrt{12.25}\Rightarrow\sqrt{12}< 3.5\)
\(20< 20.25\Rightarrow\sqrt{20}< \sqrt{20.25}\Rightarrow\sqrt{20}< 4.5\)
\(30< 30.25\Rightarrow\sqrt{30}< \sqrt{30.25}\Rightarrow\sqrt{30}< 5.5\)
\(42< 42.25\Rightarrow\sqrt{42}< \sqrt{42.25}\Rightarrow\sqrt{42}< 6.5\)
\(50< 56.5\Rightarrow\sqrt{50}< \sqrt{56.25}\Rightarrow\sqrt{50}< 7.5\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 2.5+3.5+4.5+5.5+6.5+7.5\)
\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\) \(\left(ĐPCM\right)\)
Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)
3 tháng 7 2017
\(\)\(\text{a) }\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)
Ta có : \(1< 9\Rightarrow\sqrt{1}< \sqrt{9}\Rightarrow\sqrt{1}< 3\)
\(2< 9\Rightarrow\sqrt{2}< \sqrt{9}\Rightarrow\sqrt{2}< 3\)
\(3< 9\Rightarrow\sqrt{3}< \sqrt{9}\Rightarrow\sqrt{3}< 3\)
\(...\)
\(8< 9\Rightarrow\sqrt{8}< \sqrt{9}\Rightarrow\sqrt{8}< 3\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3+3+...+3_{\left(\text{8 số hạng 3}\right)}\) \(\) \(\)
\(\) \(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 3\cdot8\)
\(\Rightarrow\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\) \(\left(ĐPCM\right)\)
Vậy \(\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)
\(\text{b) }\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)
Ta có : \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{1}}< \dfrac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\dfrac{1}{\sqrt{2}}< \dfrac{1}{\sqrt{100}}\)
\(...\)
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{100}}+...+\dfrac{1}{\sqrt{100}}_{\left(\text{100 số hạng}\dfrac{1}{\sqrt{100}}\right)}\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}\cdot100\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>\dfrac{10}{\sqrt{100}}\)
\(\Rightarrow\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\) \(\left(ĐPCM\right)\)
Vậy \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)
\(\)
22 tháng 8 2021
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
17 tháng 12 2021
\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
28 tháng 7 2022
b: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
=-2
c: \(=\dfrac{\sqrt{4}\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{\sqrt{6}}{2}\)
21 tháng 7 2018
\(a.\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=2\sqrt{5}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}-2}{1-\sqrt{5}}=\dfrac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\) \(b.\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6.5}+\sqrt{27.6}}=\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{17}\right)}=-\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-2-1}{\sqrt{6}}=-\dfrac{\sqrt{3}}{\sqrt{2}}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
\(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right):\dfrac{1}{2\sqrt{5}-\sqrt{6}}\)
\(=\left(\dfrac{2\sqrt{5}\cdot\sqrt{5}+\sqrt{2}\cdot2\sqrt{5}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{5}\cdot\sqrt{6}-\sqrt{6}\cdot1}{\sqrt{5}-1}\right):\dfrac{1}{2\sqrt{5}-\sqrt{6}}\)
\(=\left[\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right]\cdot\left(2\sqrt{5}-\sqrt{6}\right)\)
\(=\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)\)
\(=\left(2\sqrt{5}\right)^2-\left(\sqrt{6}\right)^2\)
\(=20-6\)
\(=14\)
31 tháng 8 2023
\(=\left(\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2\sqrt{5}-\sqrt{6}\right)\)
\(=\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)\)
=20-6
=14
Lời giải:
\(=\sqrt{106-2\sqrt{1440}}+3\sqrt{10}-\frac{12(\sqrt{10}-2)}{(\sqrt{10}+2)(\sqrt{10}-2)}\)
\(=\sqrt{(\sqrt{90}-\sqrt{16})^2}+3\sqrt{10}-\frac{12(\sqrt{10}-2)}{6}\)
$=\sqrt{90}-\sqrt{16}+3\sqrt{10}-2(\sqrt{10}-2)$
$=4\sqrt{10}$