Cho M=\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{100}}\)So sanh M voi 10 ta duoc
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so sanh B=\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)voi \(\frac{2}{7}\)
Cho \(M=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.........+\frac{1}{\sqrt{100}}\). So sánh \(M\)với \(10\)ta được
M<10 dung do minh vua lam xong
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Ta có:
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(....................\)
\(\frac{1}{\sqrt{98}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\).Cộng theo vế ta có:
\(M=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>\frac{99}{\sqrt{100}}=\frac{99}{10}\)
\(\Rightarrow M=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{99}{10}+\frac{1}{10}=\frac{100}{10}=10\)(M>10)
ps:tin mk đi đừng tin mấy thg chuyên đi spam copy, vv... r` hack
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Cho \(M=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}.\)So sánh M với 10
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{10}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{10}};...;\frac{1}{\sqrt{9}}>\frac{1}{\sqrt{10}};\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}\)
=>M>10
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Xem thêm câu trả lời
a) So sanh: \(\sqrt{17}+\sqrt{26}+1\)va \(\sqrt{99}\)
b) CMR: \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
a)\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\)
b) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+.......+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=10\)
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Cho A = \(\left(\frac{x-\sqrt{x}+7}{x-4}+\frac{1}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x-4}\right)\)
a) Rut gon A
b) So sanh A voi \(\frac{1}{A}\)
cho P=\(\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\\ \)
a, rut gon P
b, tinh gia tri cua P khi \(x=\frac{3-2\sqrt{2}}{4}\)
c, so sanh P voi \(\frac{3}{2}\)
\(\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{x+9}+3}+\frac{1}{\sqrt{x+16}+4}=\frac{1}{\sqrt{x+100}+10}\)
tìm x=?
Bằng 1 phép so sánh đơn giản \(\frac{1}{\sqrt{x+1}+1}>\frac{1}{\sqrt{x+100}+10}\) ; \(\forall x\ge-1\)
Ta suy ra luôn pt này vô nghiệm
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Cho A = \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
So sánh A với 10.
So sánh \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}...+\frac{1}{\sqrt{100}}\) và 10
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 \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
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