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MN
4 tháng 10 2018
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
\(S< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)
13 tháng 10 2017
ta có :
\(S=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{25}}\)\(\Leftrightarrow S=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+\dfrac{2}{2\sqrt{4}}+...+\dfrac{2}{2\sqrt{25}}\)
\(>\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+\dfrac{2}{\sqrt{4}+\sqrt{5}}+...+\dfrac{2}{\sqrt{25}+\sqrt{26}}\)
\(\Leftrightarrow S>2.\left(\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+...+\dfrac{\sqrt{25}-\sqrt{26}}{25-26}\right)\)\(\Leftrightarrow S>2\left(\dfrac{\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+\sqrt{4}-\sqrt{5}+...+\sqrt{25}-\sqrt{26}}{-1}\right)\)
\(\Leftrightarrow S>2\left(\dfrac{\sqrt{2}-\sqrt{26}}{-1}\right)\)
\(\Leftrightarrow S>2.\left(\sqrt{26}-\sqrt{2}\right)\)
\(\Leftrightarrow S>7,4\)
\(\Leftrightarrow S>7\)
chúc bn hc tốt
13 tháng 10 2017
\(Áp\:dụng\:công\:thức:\:2\left(\sqrt{k+1}-\sqrt{k}\right)< \dfrac{1}{\sqrt{k}}\:ta\:có:\\• 2\left(\sqrt{2+1}-\sqrt{2}\right)< \dfrac{1}{\sqrt{2}}\\ •2\left(\sqrt{3+1}-\sqrt{3}\right)< \dfrac{1}{\sqrt{3}}\\ .......\\ •2\left(\sqrt{25+1}-\sqrt{25}\right)< \dfrac{1}{\sqrt{25}}\)
\(cộng\: vế\: theo\: vế,\: ta\: được:\\ 2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{26}-\sqrt{25}\right)< A\\ \Leftrightarrow2.\left(\sqrt{26}-\sqrt{2}\right)< A\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \left(1\right)\)
ta có:
\(\left\{{}\begin{matrix}\sqrt{26}>\sqrt{25}\\\sqrt{2}< 1,5\end{matrix}\right.\Rightarrow\sqrt{26}-\sqrt{2}>\sqrt{25}-1,5=5-1,5=3,5\\ \Rightarrow2\left(\sqrt{26}-\sqrt{2}\right)>2.3,5=7\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \left(2\right)\)
từ (1) và (2) => A>7 (đpcm)
13 tháng 11 2018
1
a,\(\sqrt{\dfrac{36}{121}}=\sqrt{\dfrac{6^2}{11^2}}=\dfrac{6}{11}\)
\(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}=\sqrt{\dfrac{81}{100}}=\sqrt{\dfrac{9^2}{10^2}}=\dfrac{9}{10}\)
HS
26 tháng 7 2018
\(a,2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}.\sqrt{\dfrac{75}{16}}\)
\(\Leftrightarrow2.\dfrac{\sqrt{27}}{2}-\sqrt{\dfrac{48}{3}}-\dfrac{2}{5}.\dfrac{\sqrt{75}}{4}\)
\(\Leftrightarrow\sqrt{27}-\dfrac{4\sqrt{3}}{3}-\dfrac{1}{5}.\dfrac{5\sqrt{3}}{2}\)
\(\Leftrightarrow3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\dfrac{7\sqrt{3}}{6}\)
HS
26 tháng 7 2018
\(b,\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right).\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\Leftrightarrow\)\(\left[1+\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{-4}\right].\left[\dfrac{\left(5+\sqrt{5}\right).\left(1-\sqrt{5}\right)}{-4}+1\right]\)
\(\Leftrightarrow\)\(\left(1+\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{-4}\right).\left(\dfrac{5-5\sqrt{5}+\sqrt{5}-5}{-4}+1\right)\)
\(\Leftrightarrow\)\(\left(1+\dfrac{4\sqrt{5}}{-4}\right)\left(\dfrac{-4\sqrt{5}}{-4}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right).\left(1+\sqrt{5}\right)\)
<=> 1-5
=-4
MN
4 tháng 10 2018
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
6 tháng 10 2022
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
MN
30 tháng 9 2018
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
S\(< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)\)\(=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)(dpcm)
1 tháng 9 2018
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
1 tháng 9 2018
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
PA
31 tháng 7 2017
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
`@` `\text {Ans}`
`\downarrow`
\(S=\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)
`S=0,7 + 1/3 - 5/2`
`S=31/30 - 5/2 = -22/15`