vd:n=-0,8 thì sai

Chứng minh 

\(\sqrt[3]{\left(n+1\right)^2}-\sqrt[3]{n^2}< \frac{2}{3\sqrt[3]{n}}\)

\(\Leftrightarrow3\sqrt[3]{n\left(n+1\right)^2}< 2+3n\)

Lập phương 2 vế rồi rút gọn được

\(\Leftrightarrow9n+8>0\)

Đúng với mọi n dương. Ta có ĐPCM.

Cái còn lại tương tự

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

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

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)

Chứng minh gì vậy bạn??/

A=1^3+2^3+...+n^3

=(1+2+3+...+n)^2

=[n(n+1)/2]^2

=>\(\sqrt{A}=\sqrt{\left[\dfrac{n\left(n+1\right)}{2}\right]^2}=\dfrac{n\left(n+1\right)}{2}\) 

Vì n(n+1) chia hết cho 2

nên n(n+1)/2 là số tự nhiên

=>\(\sqrt{A}\) là số tự nhiên

\(A=\sqrt{4+\sqrt{4+\sqrt{4}+...}}\\ \)>0

a)

\(A=\sqrt{4+A}\Leftrightarrow A^2=4+A\Leftrightarrow A^2-A-4=0\)

\(\Delta=1+16=17\)

\(A_1=\dfrac{1+\sqrt{17}}{2}< \dfrac{1+5}{2}=3\)

\(A_2=\dfrac{1-\sqrt{17}}{2}\)<0 loại

Vậy A < 3

b) Chứng minh quy nạp

(1³+2³+.....+n³)=(1+2+3+...+n)²=> KL

b).đặt \(A=\sqrt{1^3+2^3+3^3+...+n^3}\)

ta có hằng đẳng thức: \(x^3-x=\left(x-1\right)x\left(x+1\right)\)

\(1^3+2^3+3^3+...+n^3=1^3-1+2^3-2+3^3-3+...+n^3-n+\left(1+2+3+...+n\right)\)\(=0+1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)+\dfrac{n\left(n+1\right)}{2}\)(*)

Xét \(B=1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)

\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

từ (*): \(1^3+2^3+...+n^3=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}+\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)}{2}\left[\dfrac{\left(n-1\right)\left(n+2\right)}{2}+1\right]=\dfrac{n\left(n+1\right)}{2}.\dfrac{n^2+n-2+2}{2}=\left[\dfrac{n\left(n+1\right)}{2}\right]^2\)

do đó \(A=\sqrt{\left[\dfrac{n\left(n+1\right)}{2}\right]^2}=\dfrac{n\left(n+1\right)}{2}=1+2+...+n\)(đpcm)

Ta có : \(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\frac{1}{k\left(k+1\right)}\right)=\sqrt{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\sqrt{k}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+}}\right)\)

\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)

Áp dụng : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+}}< 2\)

Vậy ta có điều phải chứng minh.