Ta đi chứng minh công thức tổng quát: \(f\left(n\right)=\frac{2n+1+\sqrt{n\left(n+1\right)}}{\sqrt{n}+\sqrt{n+1}}=\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\)

Thật vậy: \(\left[\left(n+1\right)\sqrt{n+1}-n\sqrt{n}\right]\left(\sqrt{n}+\sqrt{n+1}\right)=\left(n+1\right)\sqrt{n\left(n+1\right)}-n^2+\left(n+1\right)^2-n\sqrt{n\left(n+1\right)}=2n+1+\sqrt{n\left(n+1\right)}\)Áp dụng, ta được: \(f\left(1\right)+f\left(2\right)+...+f\left(2020\right)=\left(2\sqrt{2}-1\sqrt{1}\right)+\left(3\sqrt{3}-2\sqrt{2}\right)+\left(4\sqrt{4}-3\sqrt{3}\right)+...+\left(2021\sqrt{2021}-2020\sqrt{2020}\right)=2021\sqrt{2021}-1\)

\(f\left(n\right)=\dfrac{2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}}{\sqrt{2n+1}+\sqrt{2n-1}}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}-\sqrt{2n-1}\right)\left(2n-1+2n+1+\sqrt{\left(2n+1\right)\left(2n+1\right)}\right)}{2n+1-2n+1}\\ f\left(n\right)=\dfrac{\left(\sqrt{2n+1}\right)^3-\left(\sqrt{2n+1}\right)^3}{2}=\dfrac{\left(2n+1\right)\sqrt{2n+1}-\left(2n-1\right)\sqrt{2n+1}}{2}\)

\(\Leftrightarrow f\left(1\right)+f\left(2\right)+...+f\left(40\right)=\dfrac{3\sqrt{3}-1\sqrt{1}+5\sqrt{5}-3\sqrt{3}+...+81\sqrt{81}-79\sqrt{79}}{2}\\ =\dfrac{81\sqrt{81}-1\sqrt{1}}{2}=\dfrac{9^3-1}{2}=364\)

cần gấp ko bn 

có bạn. mai mk faj nộp r

a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả

b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)

c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)

d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)

e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)

f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)

g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)

\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)

\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)

a) lim \(\left(-3n^3+n^2-1\right)\)

minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n³ +2n

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

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

\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

\(\frac{1}{\sqrt[3]{2}}>\frac{1}{\sqrt[3]{3}}>...>\frac{1}{\sqrt[3]{n}}\)

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

Mà \(\lim\limits\frac{n-1}{\sqrt[3]{n}\left(n^2+1\right)}=\lim\limits\frac{n-1}{\sqrt[3]{n}\left(n^2+1\right)}=0\)

\(\Rightarrow\lim\limits\frac{f\left(n\right)}{n^2+1}=0\)

Bài 1

Ta có \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{\left(1+\frac{1}{2}-\frac{1}{3}\right)^2}\)

Tương tự như trên ta được

S = 1+1/2-1/3+1+1/3-1/4+...+1+1/99-1/100

   = 98 + 1/2 - 1/100

   = 9849/100