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

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

a: \(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{2+\sqrt{3}}{2}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{2-\sqrt{3}}{2}}\right)\)

\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\)

\(=\dfrac{2+\sqrt{3}}{2}:\left(1+\dfrac{\sqrt{3}+1}{2}\right)+\dfrac{2-\sqrt{3}}{2}:\left(1-\dfrac{\sqrt{3}-1}{2}\right)\)

\(=\dfrac{2+\sqrt{3}}{2}\cdot\dfrac{2}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2}\cdot\dfrac{2}{2-\sqrt{3}+1}\)

\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)

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

\(=\dfrac{6}{6}=1\)

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

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{\sqrt{4}-\sqrt{3}}{4-3}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=10-1=9\) Vậy , biểu thức A có giá trị nguyên .

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

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

Do đó:

\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)