Xét vế trái : \(\left(\sqrt{n+1}-\sqrt{n}\right)^2=2n+1-2\sqrt{n}.\sqrt{n+1}\)

Xét vế phải : \(\sqrt{\left(2n+1\right)^2}-\sqrt{\left(2n+1\right)^2-1}=\left|2n+1\right|-\sqrt{\left(2n+1-1\right)\left(2n+1+1\right)}\)

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

=> VT = VP => đpcm

Xét vế trái : \(\left(\sqrt{n+1}-\sqrt{n}\right)^2=2n+1-2\sqrt{n}.\sqrt{n+1}\)

Xét vế phải : \(\sqrt{\left(2n+1\right)^2}-\sqrt{\left(2n+1\right)^2-1}=\left|2n+1\right|-\sqrt{\left(2n+1-1\right)\left(2n+1+1\right)}=2n+1-2\sqrt{n}.\sqrt{n+1}\)

=> VT = VP 

=> đpcm

khó quá ms đầu năm s học cao thế bạn ơi 

Ta co:

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

Ap vào bài toan được

\(S_n=\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)

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

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

iopdtg5 r4ytr'hfgo;hrt687y5t53434]\trvf;lkg

Ta có: \(VT=\sqrt{\left(2n+1\right)^2}+\sqrt{4n^2}=\sqrt{\left(2n+1\right)^2}+\sqrt{\left(2n\right)^2}\)

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

Vì \(n\inℕ\)\(\Rightarrow2n+1>0\); \(2n\ge0\)

\(\Rightarrow\left|2n+1\right|=2n+1\)và \(\left|2n\right|=2n\)

\(\Rightarrow VT=2n+1+2n=4n+1\)

Ta có: \(VP=\left(2n+1\right)^2-4n^2=\left(2n+1\right)^2-\left(2n\right)^2\)

\(=\left(2n+1-2n\right)\left(2n+1+2n\right)=4n+1\)

\(\Rightarrow VT=VP\)\(\Rightarrowđpcm\)

Ở câu a ko có chữ " b " nhé

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