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

\(C=\frac{a}{a-1}\cdot\frac{a-1}{\sqrt{a}+1}\)

\(C=\frac{a}{\sqrt{a}+1}\)

Vậy ...

search google là xong mà chị

Xét hình bình hành \(ABCD\)có \(O\)là giao điểm của \(AC\)và \(BD\).

Khi đó \(O\)là trung điểm của \(AC\)và \(BD\).

Độ dài hai đường chéo tỉ lệ với độ dài hai cạnh liên tiếp nên \(\frac{BD}{AC}=\frac{AB}{AD}\Leftrightarrow\frac{DA}{OA}=\frac{AB}{OB}\).

Xét tam giác \(DAB\)và tam giác \(AOB\)có: 

\(\widehat{DBA}=\widehat{ABO}\)(góc chung) 

\(\frac{DA}{AO}=\frac{AB}{OB}\)(cmt)

Suy ra \(\Delta DAB~\Delta AOB\left(c.g.c\right)\).

suy ra \(\widehat{AOB}=\widehat{DAB}\)(hai góc tương ứng) 

Ta có đpcm.

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

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

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

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

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

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

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