\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

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

ĐK : a >= 0 , a khác 1

\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)

Xét \(A=\frac{1}{\sqrt{x^2+1}-x}=\frac{\sqrt{x^2+1}+x}{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}=\frac{\sqrt{x^2+1}+x}{x^2+1-x^2}=\sqrt{x^2+1}+x\)

Lưu ý: ĐKXĐ của A là \(x\in R\)vì \(\hept{\begin{cases}x^2+1>0\\\sqrt{x^2+1}>x\end{cases},\forall x\in R}\)

Vậy để \(A\in N\)thì \(\sqrt{x^2+1}+x=k,k\in N,k>0\Rightarrow\sqrt{x^2+1}=k-x\)

\(\Rightarrow x^2+1=x^2-2kx+k^2\Rightarrow x=\frac{k^2-1}{2k},k\in N,k>0\)

Vậy yêu cầu bài toán thỏa mãn khi x có dạng \(\frac{k^2-1}{2k},k\inℕ^∗\)

Chứng minh bài toán phụ: 

Nếu \(a+b+c=0\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) với a,b,c khác 0

Ta có: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{a+b+c}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

=> đpcm

Áp dụng vào:

a) Ta có: \(U_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{1}{1}+\frac{1}{n^2}+\frac{1}{\left[-\left(n+1\right)\right]^2}}\)

\(=\left|1+\frac{1}{n}-\frac{1}{n+1}\right|\) là số hữu tỉ vì n là số tự nhiên khác 0

b) Áp dụng công thức tự tính ra nhé

a) \(u_n=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)

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

\(=\sqrt{\frac{\left[n\left(n+1\right)+1\right]^2}{\left[n\left(n+1\right)\right]^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}\in Q\)

b) \(u_n=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)

Vậy \(S_{2021}=u_1+u_2+...+u_{2021}=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2021}-\frac{1}{2022}\)

\(=2022-\frac{1}{2022}=\frac{2022^2-1}{2022}\)

\(\sqrt{x^2+1}∈N\Leftrightarrow x^2+1=k^2\Leftrightarrow x^2=k^2-1\Leftrightarrow x^2\text{ và }k^2\text{là hai số chính phương liên tiếp }\)

\(\Leftrightarrow\hept{\begin{cases}x^2 = 0\\k^2 = 1\end{cases}\Leftrightarrow\hept{\begin{cases}x = 0\\k= 1\end{cases}}}\)

\(\sqrt{x^2+1}∈N\Leftrightarrow x^2+1=k^2\Leftrightarrow x^2=k^2-1\Leftrightarrow x^2\text{ và }k^2\text{là hai số chính phương liên tiếp }\)

\(\Leftrightarrow\hept{\begin{cases}x^2=0\\k^2=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\k=1\end{cases}}\)

Trả lời :

\(\sqrt{x^2}+1∈N\Leftrightarrow\sqrt{x^2}∈N\left(\text{do 1 ∈ N }\right)\)

√ĐKXĐ: −32x−1≥0⇔2x−1<0⇔x<12

 can 2 cong vs can may z