MC
1
K
Khách
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16 tháng 6 2017
a/ \(\sqrt{17}+\sqrt{5}+1>\sqrt{16}+\sqrt{4}+1=4+2+1=7\)
\(\sqrt{45}< \sqrt{49}=7\)
\(\Rightarrow\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)
b/ Ta có:
\(\sqrt{n}< \sqrt{n+1}\)
\(\Rightarrow2\sqrt{n}< \sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n+1}+\sqrt{n}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)
Áp dụng vào bài toán được
\(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{36}}>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{37}-\sqrt{36}\right)\)
\(=2\left(\sqrt{37}-1\right)>6\)
TN
3 tháng 9 2017
Ta có: \(M=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{224}+\sqrt{225}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{225}-\sqrt{224}\)
\(=-1+\sqrt{225}=-1+15=14\)
Và \(N=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{63}}\)
\(=14,47706...>14=M\)
QL
1
KN
8 tháng 7 2019
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
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\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100=\frac{100}{10}=10\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
22 tháng 7 2018
Ta có
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
........................................
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)(100 số\(\frac{1}{10}\)) >10
5 tháng 5 2015
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right).\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right).\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{\sqrt{121}-\sqrt{120}}{\left(\sqrt{121}-\sqrt{120}\right)\left(\sqrt{121}+\sqrt{120}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{121}-\sqrt{120}}{121-120}\)
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)
\(A=\sqrt{121}-\sqrt{1}=10\)
\(B=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+...+\frac{2}{2\sqrt{35}}\)
\(B=2.\left(\frac{1}{\sqrt{1}+\sqrt{1}}+\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{35}+\sqrt{35}}\right)\)
\(>2.\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{35}+\sqrt{36}}\right)\)
\(>2.\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\right)\)
\(=2.\left(\sqrt{36}-\sqrt{1}\right)=2.\left(6-1\right)=10=A\)
Vậy B > A
TN
2 tháng 7 2017
B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)
\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)
\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)
\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{36}}\)\(=\frac{2}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{3}}+...+\frac{2}{\sqrt{36}+\sqrt{36}}\)
\(< \frac{2}{\sqrt{1}}+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)
\(=2\left(\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)
\(=2.\sqrt{36}=12< 14\)
Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{36}}< 14\)