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\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)
\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)
a) Ta có \(\sqrt{170}>\sqrt{169}\\\)
mà \(\sqrt{169}=13\)
=> \(\sqrt{170}>13\)
b) Ta có \(\sqrt{6}< \sqrt{9}\)
mà \(\sqrt{9}=3\)
=> \(\sqrt{6}< 3\)
c) ta có \(\sqrt{226}>\sqrt{225}\)
mà \(\sqrt{225}=15\)
=>\(\sqrt{226}>15\)
d) \(\sqrt{12}>\sqrt{7}\)
e)
Ta có\(\sqrt{150}< \sqrt{180}\)
mà \(\sqrt{150}=5\sqrt{6}\)
\(\sqrt{180}=6\sqrt{5}\)
=> \(5\sqrt{6}< 6\sqrt{5}\)
\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)
\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(\sqrt{2}=\sqrt{1.2}< \dfrac{\left(1+2\right)}{2}=\dfrac{3}{2}\)
\(\sqrt{6}=\sqrt{2.3}< \dfrac{\left(2+3\right)}{2}=\dfrac{5}{2}\)
\(\sqrt{12}=\sqrt{3.4}< \dfrac{3+4}{2}=\dfrac{7}{2}\)
\(\sqrt{20}=\sqrt{4.5}< \dfrac{4+5}{2}=\dfrac{9}{2}\)
\(\sqrt{30}=\sqrt{5.6}< \dfrac{5+6}{2}=\dfrac{11}{2}\)
\(\sqrt{42}=\sqrt{6.7}< \dfrac{6+7}{2}=\dfrac{13}{2}\)
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VT \(VT=A< \dfrac{3+5+7+9+11+13}{2}=\dfrac{48}{2}=24=VP=B\)
\(\text{c) }\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{50}< 30\)
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 \(\dfrac{1}{\sqrt{10}}+\dfrac{1}{\sqrt{20}}+...\dfrac{1}{\sqrt{100}}>10\)
\(\)
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
______________________________________________
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
Ta có
\(\sqrt{2}\)=\(\sqrt{\dfrac{8}{4}}\)<\(\sqrt{\dfrac{9}{4}}\)=\(\dfrac{3}{2}\)
\(\sqrt{6}\)=\(\sqrt{\dfrac{24}{4}}\)<\(\sqrt{\dfrac{25}{4}}\)=\(\dfrac{5}{2}\)
\(\sqrt{12}\)=\(\sqrt{\dfrac{48}{4}}\)<\(\sqrt{\dfrac{49}{4}}\)=\(\dfrac{7}{2}\)
\(\sqrt{20}\)=\(\sqrt{\dfrac{80}{4}}\)<\(\sqrt{\dfrac{81}{4}}\)=\(\dfrac{9}{4}\)
\(\sqrt{30}\)=\(\sqrt{\dfrac{120}{4}}\)<\(\sqrt{\dfrac{121}{4}}\)=\(\dfrac{11}{2}\)
\(\sqrt{42}\)=\(\sqrt{\dfrac{168}{4}}\)<\(\sqrt{\dfrac{169}{4}}\)=\(\dfrac{13}{2}\)
Do đó A<\(\dfrac{3}{2}+\dfrac{5}{2}+\dfrac{7}{2}+\dfrac{9}{2}+\dfrac{11}{2}+\dfrac{13}{2}\)=24
Vậy A<24