ĐK : u, v > 0 , u khác v

\(=\frac{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}{\sqrt{u}+\sqrt{v}}-\frac{\left(\sqrt{u}+\sqrt{v}\right)\left(u-\sqrt{uv}+v\right)}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\frac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)

\(=\frac{u-2\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}=\frac{-\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)

Bài 3 : 

a, Với \(x\ge0;x\ne1\)

\(P=\frac{3x-\sqrt{x}-8}{x+\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+2}\)

\(=\frac{3x-\sqrt{x}-8-\left(x-4\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x-\sqrt{x}-8-x+4+2\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2\sqrt{x}-3}{\sqrt{x}-1}\)

b, Ta có : \(x=\frac{4}{\sqrt{7}+\sqrt{5}}+\frac{6}{2-\sqrt{7}}+10=\frac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}+\frac{6\left(2+\sqrt{7}\right)}{-3}+10\)

\(=2\sqrt{7}-2\sqrt{5}-4-2\sqrt{7}+10=-2\sqrt{5}+6\)

\(\Rightarrow\sqrt{x}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

Thay vào P ta được : \(\frac{2\left(\sqrt{5}-1\right)-3}{\sqrt{5}-1-1}=\frac{2\sqrt{5}-5}{\sqrt{5}-2}=-\sqrt{5}\)

c, Ta có : \(\frac{2\sqrt{x}-3}{\sqrt{x}-1}\le1\Leftrightarrow\frac{2\sqrt{x}-3}{\sqrt{x}-1}-1\le0\)

\(\Leftrightarrow\frac{2\sqrt{x}-3-\sqrt{x}+1}{\sqrt{x}-1}\le0\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}-1}\le0\)

Vì \(\sqrt{x}-1>\sqrt{x}-2\)

\(\hept{\begin{cases}\sqrt{x}-2\le0\\\sqrt{x}-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le4\\x\ge1\end{cases}}\Leftrightarrow1\le x\le4\)

Kết hợp với đk vậy \(1< x\le4\)

d, Ta có : \(\frac{2\sqrt{x}-3}{\sqrt{x}-1}-\frac{3}{2}\le0\Leftrightarrow\frac{4\sqrt{x}-6-3\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}\le0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}-1\right)}\le0\) TH1 : \(\hept{\begin{cases}\sqrt{x}-3\le0\\\sqrt{x}-1\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le9\\x\ge1\end{cases}}\Leftrightarrow1< x\le9\)

TH1 : \(\hept{\begin{cases}\sqrt{x}-3\ge0\\\sqrt{x}-1\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge9\\0\le x< 1\end{cases}}\)( vô lí ) 

Ta có : \(\Delta=\left(-2m\right)^2-4\left(-2m-1\right)=4m^2+8m+4\)

Để pt có 2 nghiệm phân biệt khi \(\Delta>0\)

\(\Delta=4m^2+8m+4=4\left(m^2+2m+1\right)=4\left(m+1\right)^2>0\Rightarrow m+1>0\Leftrightarrow m>-1\)

Theo Vi et : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=2m\\x_1x_2=\frac{c}{a}=-2m-1\end{cases}}\)

Thay vào ta được : \(\sqrt{2m}+\sqrt{3+\left(-2m-1\right)}=2m+1\)

\(\Leftrightarrow\sqrt{2m}+\sqrt{-2m+2}=2m+1\)

\(\Leftrightarrow2m+2\sqrt{-4m^2+4m}-2m+2=4m^2+4m+1\)

\(\Leftrightarrow2\sqrt{-4m^2+4m}+2=4m^2+4m+1\)

\(\Leftrightarrow2\sqrt{-4m^2+4m}=4m^2+4m-1\)

bạn bình phương tìm m so sánh với đk nhé ;)) 

\(Q=\frac{2017}{x-8\sqrt{x}+2018}=\frac{2017}{\left(\sqrt{x}-4\right)^2+2002}\)

ta có \(\left(\sqrt{x}-4\right)^2\ge0\)

\(Q\le\frac{2017}{2002}\)

dấu "="  xảy ra khi \(x=16\)

\(MAX:Q=\frac{2017}{2002}\)