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\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)
\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)
Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)
\(\sqrt{2}=\sqrt{1.2}< \dfrac{\left(1+2\right)}{2}=\dfrac{3}{2}\)
\(\sqrt{6}=\sqrt{2.3}< \dfrac{\left(2+3\right)}{2}=\dfrac{5}{2}\)
\(\sqrt{12}=\sqrt{3.4}< \dfrac{3+4}{2}=\dfrac{7}{2}\)
\(\sqrt{20}=\sqrt{4.5}< \dfrac{4+5}{2}=\dfrac{9}{2}\)
\(\sqrt{30}=\sqrt{5.6}< \dfrac{5+6}{2}=\dfrac{11}{2}\)
\(\sqrt{42}=\sqrt{6.7}< \dfrac{6+7}{2}=\dfrac{13}{2}\)
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VT \(VT=A< \dfrac{3+5+7+9+11+13}{2}=\dfrac{48}{2}=24=VP=B\)
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
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
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\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
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 A=1+2+3+4+5+6=\(\sqrt{1}\)+\(\sqrt{4}\)+\(\sqrt{9}\)+\(\sqrt{16}\)+\(\sqrt{25}\)+\(\sqrt{36}\)
Ta thấy \(\sqrt{1}\)<\(\sqrt{2}\)
\(\sqrt{4}\)<\(\sqrt{6}\)
.............
\(\sqrt{36}\)<\(\sqrt{42}\)
có gì sai thì sửa nhé
=>\(\sqrt{1}\)+\(\sqrt{4}\)+\(\sqrt{9}\)+\(\sqrt{16}\)+\(\sqrt{25}\)+\(\sqrt{36}\)<\(\sqrt{2}\)+\(\sqrt{6}\)+\(\sqrt{12}\)+\(\sqrt{20}\)+\(\sqrt{30}\)+\(\sqrt{42}\)
=>B<A hay A>B
ghi de sai ban oi
\(A=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)
\(A< \sqrt{2,25}+\sqrt{6,25}+\sqrt{12,25}+\sqrt{20,25}+\sqrt{30,25}+\sqrt{42,25}=24=B\)
Vậy \(A< B\)
Chúc bạn học tốt ~