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: \(t+1> A> t\)

\(\Rightarrow \lfloor{A}\rfloor=t=\frac{n}{2}\)

ai giúp e nhanh lên đc k

\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)

bai 1

(n+1)√n=√n^3+√n>2√(n^3.n)=2n^2>2(n^2-1)=2(n-1)(n+1)

1/[(n+1)√n]<1/[2(n-1)(n+1)]=1/4.[2/(n-1)(n+1)]

A=..

n =1 yes

n>1

A<1+1/4[2/1.3+2/3.5+..+2/(n-1)(n+1)

A<1+1/4[ 2-1/(n+1)]<1+1/2<2=>dpcm

b,\(B=\sqrt{1+2014^2+\dfrac{2014^2}{2015^2}}+\dfrac{2014}{2015}\)

Ta có :\(\left(2014+1\right)^2=2014^2+1+2.2014\)

\(\Rightarrow2014^2+1=2015^2-2.2014\)

\(\Rightarrow B=\sqrt{2015^2-2.2014+\left(\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)

\(=\sqrt{\left(2015-\dfrac{2014}{2015}\right)^2}+\dfrac{2014}{2015}\)

\(=2015-\dfrac{2014}{2015}+\dfrac{2014}{2015}\)

\(=2015\)

Vậy B=2015

Một câu thôi: Liên hợp

\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)

\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)

\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(

\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)

\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)

\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)

Lim 3.4n-2.13n/5n+6.13n