1, \(x^3=\left(7+\sqrt{\frac{49}{8}}\right)+\left(7-\sqrt{\frac{49}{8}}\right)+3x\sqrt[3]{\left(7+\sqrt{\frac{49}{8}}\right)\left(7-\sqrt{\frac{49}{8}}\right)}\)

\(=14+3x\cdot\frac{7}{2}=14+\frac{21x}{2}\)

\(\Leftrightarrow x^3-\frac{21}{2}x-14=0\)

Ta có: \(f\left(x\right)=\left(2x^3-21-29\right)^{2019}=\left[2\left(x^3-\frac{21}{2}x-14\right)-1\right]^{2019}=\left(-1\right)^{2019}=-1\)

2, ta có: \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2=\left[\frac{n\left(n+1\right)}{2}\right]^2\) (bạn tự cm)

Áp dụng công thức trên ta được n=2016

3, \(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\frac{\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}\)

\(=\frac{\sqrt[3]{\left(\sqrt{5}-2\right)^3}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{\sqrt{5}+3-\sqrt{5}}=\frac{5-4}{3}=\frac{1}{3}\)

Thay x=1/3 vào A ta được;

\(A=3x^3+8x^2+2=3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2=3\)

Bài 4

ÁP DỤNG BĐT CAUCHY 

là ra

\(\frac{1}{\sqrt{1^3+2^3}}+\frac{1}{\sqrt{1^3+2^3+3^3}}+...+\frac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\frac{2015}{2017}\) (1) 

Cần CM: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\) quy nạp nhé bn, trên mạng có nhìu 

(1) \(\Leftrightarrow\)\(\frac{1}{\sqrt{\left(1+2\right)^2}}+\frac{1}{\sqrt{\left(1+2+3\right)^2}}+...+\frac{1}{\sqrt{\left(1+2+3+...+n\right)^2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+n}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+...+\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2017}\)

\(\Leftrightarrow\)\(n=2016\)

Xét hạng tổng quát:

\(\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{\sqrt{n}-\sqrt{n-1}}{n-n+1}=\sqrt{n}-\sqrt{n-1}\)

Áp dụng vào bài, ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

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

\(=\sqrt{n}-1\)

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\ge2014\)

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

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

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

\(=\frac{1-\sqrt{n+1}}{-1}=\sqrt{n+1}-1\ge2014\)

                                  \(\Leftrightarrow\sqrt{n+1}\ge2015\)

                                 \(\Leftrightarrow n+1=2015^2=4060225\)

\(V~~n=4060224\)

Xét \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{\left(n+1\right)n}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) = \(\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) < \(2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Vậy \(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.....+\frac{1}{\left(n+1\right)\sqrt{n}}<2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) = \(2\left(1-\frac{1}{\sqrt{n+1}}\right)<2\) (đpcm)

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

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

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

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

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

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

......................?

mik ko biết

mong bn thông cảm 

nha ................

\(S=\frac{1}{3\sqrt{1}+3\sqrt{3}}+\frac{1}{3\sqrt{3}+3\sqrt{5}}+...+\frac{1}{3\sqrt{2017}+3\sqrt{2019}}\)

\(S=\frac{1}{3}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{2017}+\sqrt{2019}}\right)\)

\(S=\frac{1}{3}\left[\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+...+\frac{\sqrt{2019}-\sqrt{2017}}{2019-2017}\right]\)

\(S=\frac{1}{3}\cdot\frac{\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2017}}{2}\)

\(S=\frac{\sqrt{2019}-\sqrt{1}}{6}\)

\(2S=\frac{1}{3}\left(\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{3}+\sqrt{5}}+...+\frac{2}{\sqrt{2017}+\sqrt{2019}}\right)\)

\(2S=\frac{1}{3}\left(\frac{3-1}{\sqrt{1}+\sqrt{3}}+\frac{5-3}{\sqrt{3}+\sqrt{5}}+...+\frac{2019-2017}{\sqrt{2017}+\sqrt{2019}}\right)\)

\(2S=\frac{1}{3}\left(\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{3}+\sqrt{5}}+...+\frac{\left(\sqrt{2019}-\sqrt{2017}\right)\left(\sqrt{2019}+\sqrt{2017}\right)}{\sqrt{2019}+\sqrt{2017}}\right)\)

\(2S=\frac{1}{3}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2017}\right)\)

\(S=\frac{\sqrt{2019}-1}{6}\)