√50-4√2+2√72-√32 = 12.7279220614

ta có hệ tương đương 

\(\hept{\begin{cases}\frac{1}{\sqrt{x}-1}+\frac{6}{|y-2|}=2\\\frac{2}{\sqrt{x}-1}-1-\frac{1}{|2-y|}=-9\end{cases}}\)Đặt \(\hept{\begin{cases}\frac{1}{\sqrt{x}-1}=a\\\frac{1}{\left|y-2\right|=b}\end{cases}\Rightarrow HPT\Leftrightarrow\hept{\begin{cases}a+6b=2\\2a-b=-8\end{cases}}}\)

\(\hept{\begin{cases}a=-\frac{46}{13}\\b=\frac{12}{13}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1089}{2116}\\y=\frac{37}{12}\text{ hoặc }y=\frac{14}{13}\end{cases}}}\)

omg, khó thế, me mới lớp four thôi sis

haha không sao

\(A=\frac{4\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}+8-8}{\sqrt{x}+2}=4-\frac{8}{\sqrt{x}+2}\)(x \(\ge0\))

Nhận thấy \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Rightarrow-\frac{8}{\sqrt{x}+2}\ge-4\)

<=> \(4-\frac{8}{\sqrt{x}+2}\ge0\)

Dấu "=" xảy ra <=> x = 0

Vậy Min A = 0 <=> x = 0 

Xét hiệu  \(D=\sqrt{2016}-\sqrt{2014}-\left(\sqrt{2017}-\sqrt{2016}\right)\)ta có:

\(D=\sqrt{2016}-\sqrt{2014}-\sqrt{2017}+\sqrt{2016}=2\sqrt{2016}-\sqrt{2014}-\sqrt{2017}\)

Ta thử so sánh \(2\sqrt{2016}\)và \(\sqrt{2014}+\sqrt{2017}\)

Ta có \(\left(\sqrt{2014}+\sqrt{2017}\right)^2=2014+2017+2\sqrt{2014.2017}\)

\(=4031+2\sqrt{\left(2015,5-1,5\right)\left(2015,5+1,5\right)}=4031+2\sqrt{\left(2015,5\right)^2-\left(1,5\right)^2}\)

Mặt khác \(\left(2\sqrt{2016}\right)^2=4.2016=2.2016+2.2016=4032+2\sqrt{2016^2}\)

Ta thấy 4032 > 4031 và 2016² > (2015,5)² - (1,5)² (hiển nhiên) 

\(\Rightarrow4032+2\sqrt{2016^2}>4031+2\sqrt{\left(2015,5\right)^2-\left(1,5\right)^2}\)

\(\Leftrightarrow\left(2\sqrt{2016}\right)^2>\left(\sqrt{2014}+\sqrt{2017}\right)^2\)

\(\Leftrightarrow2\sqrt{2016}>\sqrt{2014}+\sqrt{2017}\)

\(\Rightarrow D>0\)

\(\Rightarrow\sqrt{2016}-\sqrt{2014}>\sqrt{2017}-\sqrt{2016}\)