\(\sqrt{n}< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n},\forall n\ge1\)
Chứng minh: \(\sqrt{n}< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n},\forall n\ge1\)
Chứng minh với \(\forall\)n \(\in\)N có
\(\dfrac{1}{2\sqrt{2}+1}+\dfrac{1}{3\sqrt{3}+2\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< 1-\dfrac{1}{\sqrt{n+1}}\)
Bạn xem lời giải tại đây:
cmr:
\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(\forall n\ge1\right)\)
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)
CMR, ∀n ≥ 1, n ∈ N : \(\dfrac{1}{2}\)+\(\dfrac{1}{3\sqrt{2}}\)+\(\dfrac{1}{4\sqrt{3}}\)+....+ \(\dfrac{1}{\left(n+1\right)\sqrt{n}}\)<2
CMR, ∀n ≥ 1, n ∈ N : \(\dfrac{1}{2}\)+\(\dfrac{1}{3\sqrt{2}}\)+\(\dfrac{1}{4\sqrt{3}}\)+....+ \(\dfrac{1}{\left(n+1\right)\sqrt{n}}\)<2
Rút gọn
\(A=\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}+\dfrac{1}{\sqrt{2}-\sqrt{3}}+....+\dfrac{1}{\sqrt{n-1}-\sqrt{n}}\) (n thuộc N, n>=2)
trong bai :
cho a= \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+....+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)
co phan huong dan : \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
cho minh hoi buoc : \(\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\) tu dau ra .( giai thich chi tiet)
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
a)tính tổng S=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+..+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\)
b)Áp dụng, tìm phần nguyên của A=\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+...+\dfrac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\) với n lẻ
Câu a : Ta có :
\(\dfrac{1}{1+\sqrt{2}}=\dfrac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}=\dfrac{1-\sqrt{2}}{1-2}=\dfrac{1-\sqrt{2}}{-1}=-1+\sqrt{2}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{2}-\sqrt{3}}{2-3}=\dfrac{\sqrt{2}-\sqrt{3}}{-1}=-\sqrt{2}+\sqrt{3}\)
.....................
\(\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}=\dfrac{\sqrt{n^2-1}-\sqrt{n^2}}{\left(\sqrt{n^2-1}+\sqrt{n^2}\right)\left(\sqrt{n^2-1}-\sqrt{n^2}\right)}=\dfrac{\sqrt{n^2-1}-\sqrt{n^2}}{-1}=-\sqrt{n^2-1}+\sqrt{n^2}\)
Thay vào ta được :
\(S=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n^2-1}+\sqrt{n^2}}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...........-\sqrt{n^2-1}+\sqrt{n^2}\)
\(=-1+\sqrt{n^2}\)
Câu b:
Đặt biểu thức đã cho là $A$
Ta có:
\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)
\(\Leftrightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n^2-1}+\sqrt{n^2}}\right)\)
\(\Leftrightarrow A> \frac{1}{2}(n-1)\) (áp dụng cách tính toán phần a)
Lại có:
\(A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{1+\sqrt{2}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\right)+....+\frac{1}{2}\left(\frac{1}{\sqrt{n^2-3}+\sqrt{n^2-2}}+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)
\(\Leftrightarrow A< \frac{1}{2}\left(\frac{1}{0+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{n^2-2}+\sqrt{n^2-1}}\right)\)
\(\Leftrightarrow A< \frac{\sqrt{n^2-1}}{2}\) (áp dụng cách tính toán của phần a)
Vậy \(\frac{\sqrt{n^2-1}}{2}> A> \frac{n-1}{2}\) hay \(\sqrt{t(t+1)}> A> t\) (đặt \(n=2t+1\) )
Mà \(\sqrt{t(t+1)}< \sqrt{(t+1)(t+1)}=t+1\)
Do đó: \(t+1> A> t\)
\(\Rightarrow \lfloor{A}\rfloor=t=\frac{n}{2}\)