a/ \(\frac{56}{9}\)

b/ \(1296\)

a) \(\frac{\sqrt{640}\sqrt{34,3}}{\sqrt{567}}\)

\(= \frac{\sqrt{64.10}\sqrt{49.\frac{7}{10}}}{\sqrt{81.7}}\)

\(= \frac{\sqrt{64}\sqrt{10}\sqrt{49}\sqrt{\frac{7}{10}}}{\sqrt{81}\sqrt{7}}\)

\(= \frac{\sqrt{64}\sqrt{49}}{\sqrt{81}} . \frac{\sqrt{10}\sqrt{\frac{7}{10}}}{\sqrt{7}}\)

\(= \frac{8.7}{9} . \frac{\sqrt{10 . \frac{7}{10}}}{\sqrt{7}}\)

\(= \frac{56}{9} . \frac{\sqrt{7}}{\sqrt{7}}\)

\(= \frac{56}{9} . 1 = \frac{56}{9}\)

b) \(\sqrt{21,6}\sqrt{810}\sqrt{11^2−5^2}\)

\(= \sqrt{216.\frac{1}{10}}\sqrt{81.10}\sqrt{(11−5)(11+5)}\)

\(= \sqrt{36.6.\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6.16}\)

\(= \sqrt{36}\sqrt{6}\sqrt{\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6}\sqrt{16}\)

\(= (\sqrt{36}\sqrt{81}\sqrt{16}).(\sqrt{6}\sqrt{6}).(\sqrt{\frac{1}{10}}\sqrt{10})\)

\(= (6.9.4).\sqrt{6.6}.\sqrt{\frac{1}{10}.10}\)

\(= (54.4).\sqrt{36}.\sqrt{1}\)

\(= 216.6.1 = 1296\)

ta tính VT ra xong rồi nói VT = VP

a)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

\(\Leftrightarrow\sqrt{1}=1\) (đpcm)

\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}-1=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{3}}-1=0\)

\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)-1=0\)

\(\Leftrightarrow\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2-1=0\)

\(\Leftrightarrow3-2-1=0\) (đpcm)

Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\) 
⇔A=  \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
 

\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)

=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)

=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)

=\(\sqrt[3]{\dfrac{125}{27}}\)

=5/3

\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

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 \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)

\(\)\(\text{a) }\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{8}< 24\)

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 \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)

\(\)

Cho phép mình chữa đề câu \(c\) thành như thế này nhé Fairy Tail

\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\)

Ta có : \(6< 9\Rightarrow\sqrt{6}< \sqrt{9}\Rightarrow\sqrt{6}< 3\)

\(12< 16\Rightarrow\sqrt{12}< \sqrt{16}\Rightarrow\sqrt{12}< 4\)

\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Rightarrow\sqrt{20}< 5\)

\(30< 36\Rightarrow\sqrt{30}< \sqrt{36}\Rightarrow\sqrt{30}< 6\)

\(42< 49\Rightarrow\sqrt{42}< \sqrt{49}\Rightarrow\sqrt{42}< 7\)

\(50< 64\Rightarrow\sqrt{50}< \sqrt{64}\Rightarrow\sqrt{50}< 8\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

\(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 3+4+5+6+7+8\)

\(\Rightarrow\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\) \(\left(ĐPCM\right)\)

Vậy \(\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 33\)

a) \(\sqrt{\dfrac{25}{81}.\dfrac{16}{49}.\dfrac{196}{9}}=\sqrt{\dfrac{25}{81}}.\sqrt{\dfrac{16}{49}}.\sqrt{\dfrac{196}{9}}=\dfrac{5}{9}.\dfrac{4}{7}.\dfrac{14}{3}=\dfrac{40}{27}\)

b) \(\sqrt{3\dfrac{1}{16}.2\dfrac{14}{25}.2\dfrac{34}{81}}=\sqrt{\dfrac{49}{16}.\dfrac{64}{25}.\dfrac{196}{81}}=\sqrt{\dfrac{49}{16}}.\sqrt{\dfrac{64}{25}}.\sqrt{\dfrac{196}{81}}=\dfrac{7}{4}.\dfrac{8}{5}.\dfrac{14}{9}=\dfrac{196}{45}\)

c) \(\dfrac{\sqrt{640}.\sqrt{34,3}}{\sqrt{567}}=\sqrt{\dfrac{640.34,3}{567}}=\sqrt{\dfrac{64.49}{81}}=\dfrac{\sqrt{64}.\sqrt{49}}{\sqrt{81}}=\dfrac{8.7}{9}=\dfrac{56}{9}\)

d) \(\sqrt{21,6}.\sqrt{810}.\sqrt{11^2-5^2}=\sqrt{21,6.810.\left(11^2-5^2\right)}=\sqrt{216.81.\left(11+5\right)\left(11-5\right)}=\sqrt{36^2.9^2.4^2}=36.9.4=1296\)