Tìm đk , rút gọn

ĐK : x > 2 

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

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

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

Với x > 2 

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

a, Với \(x\ge0;x\ne\frac{16}{9};4\)

\(P=\frac{2\sqrt{x}-4}{3\sqrt{x}-4}-\frac{4+2\sqrt{x}}{\sqrt{x}-2}+\frac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

\(=\frac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

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

b, \(P\ge-\frac{3}{4}\Rightarrow\frac{\sqrt{x}+1}{2-\sqrt{x}}+\frac{3}{4}\ge0\Leftrightarrow\frac{4\sqrt{x}+4+6-3\sqrt{x}}{8-4\sqrt{x}}\ge0\Leftrightarrow\frac{\sqrt{x}+10}{8-4\sqrt{x}}\ge0\)

\(\Rightarrow2-\sqrt{x}\ge0\Leftrightarrow x\le4\)Kết hợp với đk vậy \(0\le x< 4\)

a)\(x-4\ne0;x\ge0\)

<=>\(x\ne4;x\ge0\)

b)\(B=\left(\frac{1}{\sqrt{x}+2}-\frac{2}{x+4\sqrt{x}+4}\right):\left(\frac{2}{x-4}-\frac{1}{\sqrt{x}-2}\right)\)

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

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

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

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

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

ta có :

\(P=\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(\frac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}=\frac{6x+4\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(=\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\)

Để \(P\ge-6\Leftrightarrow\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\ge-6\Leftrightarrow\frac{48\sqrt{x}-4}{7\sqrt{x}-1}\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}0\le\sqrt{x}\le\frac{1}{12}\\\sqrt{x}>\frac{1}{7}\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x\le\frac{1}{144}\\x>\frac{1}{49}\end{cases}}\)