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a, sửa đề 

ĐK : x >= 0  \(\sqrt{4x-2\sqrt{x}+5}=\sqrt{4x-2.2\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+5}\)

\(=\sqrt{\left(2\sqrt{x}-\frac{1}{2}\right)^2+\frac{19}{4}}\ge\frac{\sqrt{19}}{2}\)

Dấu ''='' xảy ra khi \(x=\frac{1}{16}\)(tm)

Vậy GTNN của biểu thức trên bằng \(\frac{\sqrt{19}}{2}\)tại x = 1/16 

b, ĐK : 0 < x < 1 

\(\sqrt{x^2-4x+4}-\sqrt{x-2\sqrt{x}+1}\)

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

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

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

Dấu ''='' xảy ra khi x = 1/4 (tm)

Vậy GTNN của biểu thức trên bằng -5.4 tại x = 1/4 

Vậy GTNN của biểu thức trên bằng -5/4 tại x = 1/4 nhé :> 

a, \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{2-\sqrt{3}}-\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+\sqrt{3}}{2}-\frac{2+\sqrt{3}}{1}-\frac{2\left(\sqrt{5}+1\right)}{4}\)

\(=\frac{\sqrt{5}+\sqrt{3}-\sqrt{5}-1}{2}-2-\sqrt{3}=\frac{\sqrt{3}-1}{2}-\frac{4+2\sqrt{3}}{2}=\frac{-\sqrt{3}-5}{2}\)

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

\(=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{3}+3}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{3}-3}=\frac{\sqrt{2}\left(\sqrt{3}-3\right)^2+\sqrt{2}\left(\sqrt{3}+3\right)^2}{-6}\)

\(=\frac{\sqrt{2}\left(3-6\sqrt{3}+9+3+6\sqrt{3}+9\right)}{-6}=\frac{24\sqrt{2}}{-6}=-4\sqrt{2}\)

a)\(\frac{2\sqrt{x}+6}{x-9}\)=\(\frac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)=\(\frac{2}{\sqrt{x}-3}\)

    \(\frac{3\sqrt{x}-x\sqrt{3}}{\sqrt{x}-\sqrt{3}}\)=\(\frac{\sqrt{3x}\left(\sqrt{3}-\sqrt{x}\right)}{\sqrt{x}-\sqrt{3}}\)= \(-\sqrt{3x}\)

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

b)\(\frac{\sqrt{x}-3}{x-6\sqrt{x}+9}\)=\(\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)^2}\)= \(\frac{1}{\sqrt{x}-3}\)

    \(\frac{x\sqrt{x}-3\sqrt{3}}{\sqrt{x}-\sqrt{3}}\)= \(\frac{\sqrt{x}^3-\sqrt{3}^3}{\sqrt{x}-\sqrt{3}}\)= \(\frac{\left(\sqrt{x}-\sqrt{3}\right)\left(x+\sqrt{3x}+3\right)}{\left(\sqrt{x}-\sqrt{3}\right)}\)= \(x+3\sqrt{x}+3\)

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

c) \(\frac{2\sqrt{x}+2\sqrt{y}}{x-y}\) = \(\frac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)=\(\frac{2}{\sqrt{x}-\sqrt{y}}\)

     \(\frac{x\sqrt{x}+y\sqrt{y}}{x-y}\)= \(\frac{\sqrt{x}^3+\sqrt{y}^3}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)=\(\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\)

      \(\frac{x-2\sqrt{xy}+y-3}{\sqrt{x}-\sqrt{y}+\sqrt{3}}\)= \(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-\sqrt{3}^2}{\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}\)=\(\frac{\left(\sqrt{x}-\sqrt{y}-\sqrt{3}\right)\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}{\left(\sqrt{x}-\sqrt{y}+\sqrt{3}\right)}\)=\(\sqrt{x}-\sqrt{y}-\sqrt{3}\)

HaHaHaHaHaHaHaHaHAaHa

a. điều kiện : \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1\end{cases}}}\)

b, ta có : \(A=\left(\frac{\sqrt{x}-2}{\left(x+2\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{x-1}\right)\times\frac{x-1}{2\sqrt{x}}\)

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

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

c. Để A nguyên thì \(\sqrt{x}+1\) là ước của 3 hay : \(\orbr{\begin{cases}\sqrt{x}+1=1\\\sqrt{x}+1=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{ loại}\right)\\x=4\end{cases}}}\)

\(A=\left(\sqrt{x}-\frac{x+2\sqrt{x}+3}{\sqrt{x}+3}\right).\left(1+\frac{5}{\sqrt{x}-2}\right)=\left(\frac{x+3\sqrt{x}-\left(x+2\sqrt{x}+3\right)}{\sqrt{x}+3}\right).\left(\frac{\sqrt{x}-2+5}{\sqrt{x}-2}\right)\)

\(=\frac{\sqrt{x}-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)

b.ta có \(A=1-\frac{1}{\sqrt{x}-2}>1\Leftrightarrow\frac{1}{\sqrt{x}-2}< 0\Leftrightarrow0\le x< 4\)