Xét: \(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\frac{\left(n+1\right)^2+n^2\left(n+1\right)^2+n^2}{\left(n+1\right)^2}}\) (với \(n\inℕ\))

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

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

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

Áp dụng vào ta tính được: \(\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}=2015+\frac{1}{2016}+\frac{2015}{2016}\)

\(=2015+1=2016\)

Khi đó: \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2016\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2016\)

Đến đây xét tiếp các TH nhé, ez rồi:))

chẳng biết đúng ko,mới lớp 5

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2015^2+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{2x}+\sqrt{1}+\sqrt{x^2}-\sqrt{4x}+\sqrt{4}=\sqrt{1}+\sqrt{2015^2}+\sqrt{\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\sqrt{x^2}-\sqrt{6x}+3=1+2015+\frac{2015}{2016}+\frac{2015}{2016}\)

\(x-\sqrt{6x}=1+\frac{2015}{1+2016+2016}-3\)

\(x-\sqrt{6x}=2-\frac{2015}{4033}\)

\(x-\sqrt{6x}=\frac{6051}{4033}\)

pt <=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=\sqrt{1+2.2015+2015^2-2.2015+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow\left(x-1\right)+\left(x-2\right)=\sqrt{2016^2-2.2016.\frac{2015}{2016}+\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=\sqrt{\left(2016-\frac{2015}{2016}\right)^2}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016-\frac{2015}{2016}+\frac{2015}{2016}\)

\(\Leftrightarrow2x-3=2016\)

\(\Leftrightarrow2x=2019\)

\(\Leftrightarrow x=\frac{2019}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(...\right)=0\Rightarrow x=\frac{1}{2}\)

\(\frac{2x-1}{2020}-\frac{2x-1}{2019}+\frac{2x-1}{2018}=\frac{2x-1}{2017}-\frac{2x-1}{2016}\\ \Leftrightarrow\frac{2x-1}{2020}-\frac{2x-1}{2019}+\frac{2x-1}{2018}-\frac{2x-1}{2017}+\frac{2x-1}{2016}=0\\ \Leftrightarrow\left(2x-1\right)\left(\frac{1}{2020}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}+\frac{1}{2016}\right)=0\)

mà \(\frac{1}{2020}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}+\frac{1}{2016}\ne0\)

thì \(2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\frac{1}{2}\)

vậy \(x=\frac{1}{2}\)

ĐKXĐ: \(x\ne\left\{0;\frac{1}{2}\right\}\)

\(\Leftrightarrow\frac{3+x}{2x-1}-\frac{2x}{x\left(2x-1\right)}-3-\frac{x-4}{x}=0\)

\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(2x-1\right)}-\frac{2x}{x\left(2x-1\right)}-\frac{3x\left(2x-1\right)}{x\left(2x-1\right)}-\frac{\left(x-4\right)\left(2x-1\right)}{x\left(2x-1\right)}=0\)

\(\Leftrightarrow x^2+3x-2x-6x^2+3x-2x^2+9x-4=0\)

\(\Leftrightarrow-7x^2+13x-4=0\)

\(\Rightarrow x=\frac{13\pm\sqrt{57}}{14}\)

Kết quả xấu quá, chắc bạn ghi ko đúng đề

a) \(\frac{2x}{x+2}+\frac{x+2}{2x}=2\)

\(\Leftrightarrow4x^2+\left(x+2\right)^2=4x\left(x+2\right)\)

\(\Leftrightarrow5x^2+4x+4=4x^2+8x\)

\(\Leftrightarrow5x^2+4x+4-4x^2-8x=0\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow x^2-2.x.2+2^2=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Rightarrow x=2\)