ko bit

Ko biết

mik ko bít

I don't now

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.............

ko bít thì đừng comment :))

\(\forall n\inℕ^∗\)ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)  (*)

Thay n=1; n=2; n=3; .....; n=2004 Ta có:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}\)

mong các bạn giải sớm cho mk nha xin cảm ơn

Ta có

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Từ đó ta có

\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}=\frac{\sqrt{2005}-1}{\sqrt{2005}}\)

Giải:

\(A=\dfrac{2005^2-2004}{2005^3+1}\)

\(\Leftrightarrow A=\dfrac{2005^2-2005+1}{\left(2005+1\right)\left(2005^2-2005+1\right)}\)

\(\Leftrightarrow A=\dfrac{1}{2005+1}\left(1\right)\)

\(B=\dfrac{2005^2+2006}{2005^3-1}\)

\(\Leftrightarrow B=\dfrac{2005^2+2005+1}{\left(2005-1\right)\left(2005^2+2005+1\right)}\)

\(\Leftrightarrow B=\dfrac{1}{2005-1}\left(2\right)\)

Ta có:

\(\left(1\right)< \left(2\right)\)

\(\Leftrightarrow A< B\)

Vậy ...