$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{x}}<2\left(n-1\right)$
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VH
10 tháng 10 2019
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}=\frac{6}{6}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)
12 tháng 8 2019
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
12 tháng 8 2019
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
Sửa: \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{x}}< 2\left(\sqrt{x}-1\right)\)
Cần cm: \(\dfrac{1}{\sqrt{k}}< \dfrac{2}{\sqrt{k}+\sqrt{k-1}}\left(k\in N\text{*},k\ge2\right)\)
\(\Leftrightarrow\sqrt{k}+\sqrt{k-1}< 2\sqrt{k}\\ \Leftrightarrow\sqrt{k}>\sqrt{k-1}\\ \Leftrightarrow k>k-1\left(luôn.đúng\right)\)
Áp dụng: \(\dfrac{1}{\sqrt{2}}< \dfrac{2}{\sqrt{2}+\sqrt{1}};...;\dfrac{1}{\sqrt{x}}< \dfrac{2}{\sqrt{x}+\sqrt{x-1}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{x}}< \dfrac{2}{\sqrt{2}+\sqrt{1}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}+...+\dfrac{2}{\sqrt{x}+\sqrt{x-1}}\\ \Leftrightarrow\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{x}}< 2\left(\sqrt{2}-\sqrt{1}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+...+2\left(\sqrt{x}-\sqrt{x-1}\right)\\ \Leftrightarrow\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{x}}< 2\sqrt{2}-2+2\sqrt{3}-2\sqrt{2}+...+2\sqrt{x}-2\sqrt{x-1}\\ \Leftrightarrow\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{x}}< 2\sqrt{x}-2=2\left(\sqrt{x}-1\right)\)