ĐKXĐ: \(x\ge0,x\ne1\)

\(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)

= \(\dfrac{x+\sqrt{x}+1}{x+1}:\left(\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)

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

= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-1\)

= \(\dfrac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\dfrac{x+2}{\sqrt{x}-1}\)

\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}a^{1008}\)

\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}=2a^{1008}b^{1008}+2b^{1008}c^{1008}+2c^{1008}a^{1008}\)

\(\Rightarrow\left(a^{2016}-2a^{1008}b^{1008}+b^{1008}\right)+\left(b^{2016}-2b^{1008}c^{1008}+c^{1008}\right)\)\(+\left(c^{2016}-2c^{1008}a^{1008}+a^{2016}\right)=0\)

\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)=0\)

Vì \(\hept{\begin{cases}\left(a^{1008}-b^{1008}\right)^2\ge0\\\left(b^{1008}-c^{1008}\right)^2\ge0\\\left(c^{1008}-a^{1008}\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2\ge0\)

Dấu " = " xảy ra: \(\Leftrightarrow\hept{\begin{cases}a^{1008}-b^{1008}=0\\b^{1008}-c^{1008}=0\\c^{1008}-a^{1008}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a^{1008}=b^{1008}\\b^{1008}=c^{1008}\\c^{1008}=a^{1008}\end{cases}\Leftrightarrow}a=b=c\)

Thay a=b=c vào A ta có: \(A=\left(a-a\right)^{15}+\left(a-a\right)^{2015}+\left(a-a\right)^{2016}=0\)

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a khác b ta có:

\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)

\(\Rightarrow\sqrt{2016}+\sqrt{2014}< \sqrt{2015}.2\)

\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

sửa lại tí nha

Cho a,b>0 thoa mãn ab>2015a+2016b. CMR: \(a+b>\left(\sqrt{2015}+\sqrt{2016}\right)^2\)

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

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

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

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

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

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

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