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

=> a = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)\)+....+\(\frac{1}{2}\left(\frac{1}{2018}-\frac{1}{2019}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{4}\left(1-\frac{1}{2019.1010}\right)\)=\(\frac{2019.1010-1}{2.2019.2020}\)

b) tương tự \(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)=\(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)-\(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\)=\(\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)+\(\frac{1}{6}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)= M-P+N

Với n từ 1 đến 2017 thì

M= \(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{2}\right)+\frac{1}{6}\left(\frac{1}{2}-\frac{1}{3}\right)+...\)+\(\frac{1}{6}\left(\frac{1}{2017}-\frac{1}{2018}\right)\)=\(\frac{1}{6}\left(1-\frac{1}{2018}\right)=\frac{2017}{6.2018}\)

N= \(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{4}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{5}\right)+...+\)\(\frac{1}{6}\left(\frac{1}{2019}-\frac{1}{2020}\right)=\)\(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{2020}\right)=\frac{2017}{6.3.2020}\)

P= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}-\frac{1}{4}\right)+...+\)\(\frac{1}{3}\left(\frac{1}{2018}-\frac{1}{2019}\right)\)= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2019}\right)=\frac{2017}{3.2.2019}\)

M+N-P = \(\frac{2017}{6}\left(\frac{1}{2018}+\frac{1}{3.2020}-\frac{1}{2019}\right)\)=\(\frac{2017}{6}.\left(\frac{1}{2018.2019}+\frac{1}{3.2020}\right)\)

=  \(\frac{2017\left(1010+1009.673\right)}{3.2018.2019.2020}\)

Ta có \(\left(x+y\right)^5=120y+3< 120\left(x+y\right)\)

\(\Rightarrow\left(x+y\right)^4< 120< 4^4\)

\(\Rightarrow x+y< 4\)

Cho mình hỏi đề bài tìm nghiệm nguyên dương chứ nhỉ

Đề bài chỉ nói tìm nghiệm nguyên thôi

Câu 1: Sửa lạ đề chút nhé : 4x + 1  -> 4x -1 

 Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)

=>  \(\sqrt{2}.A\)= ​\(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)​

= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)

= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)

Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)

nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2

=> \(A=2:\sqrt{2}=\sqrt{2}\)

Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)

=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)

=> ​\(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)​

=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)

=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)

=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)

\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)

HPT\(\Leftrightarrow\hept{\begin{cases}x^2\left(1+y^2\right)=2\\1+x^2\left(1+y^2\right)+xy=4x^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2\left(1+y^2\right)=2\left(1\right)\\4x^2-xy=3\left(2\right)\end{cases}}\)

Xet \(x=0\)

Thay vao PT\(x^2\left(1+y^2\right)=2\)ta duoc:

\(0=2\)(khong thoa man)

Xet \(x\ne0\)

PT(2)\(\Leftrightarrow y=\frac{4x^2-3}{x}\)

Thay \(y=\frac{4x^2-3}{x}\)vao PT(1) ta duoc:

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

\(\Leftrightarrow x^2+\left(4x^2-3\right)^2=2\)

\(\Leftrightarrow16x^4-23x^2+7=0\)

Dat \(x^2=t\left(t>0\right)\)

\(\Rightarrow16t^2-23t+7=0\)

Ta co:

\(\Delta=\left(-23\right)^2-4.16.7=81>0\)

\(\Rightarrow\hept{\begin{cases}t_1=1\\t_2=\frac{7}{16}\end{cases}}\)

\(\Rightarrow x_1=1;x_2=-1;x_3=\frac{\sqrt{7}}{4};x_4=-\frac{\sqrt{7}}{4}\)

\(\Rightarrow y_1=1;y_2=1;y_3=-\frac{5\sqrt{7}}{7};y_4=\frac{5\sqrt{7}}{7}\)

Vay nghiem cua HPT la \(\left(1;1\right),\left(-1;1\right),\left(\frac{\sqrt{7}}{4};-\frac{5\sqrt{7}}{7}\right),\left(-\frac{\sqrt{7}}{4};\frac{5\sqrt{7}}{7}\right)\)

Do A thuộc đường tròn dk BC -> AB vuông góc với AC

Ta có: BAH và ACI cùng phụ với ABC -> BAH = ACI (1)

Dễ dàng CM dc tam giác ABC đồng dạng với tam giác HAC -> AB/AH = AC/HC -> AB.CH = AH.AC <=> (2.AB.)(1/2.CH) = AH.AC

<=> AM.CI = AH.AC <=> AM/AH = AC/CI (2)

Từ (1),(2) -> Tam giác AHM đồng dạng tam giác CIA