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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 \(P\ge-6\Leftrightarrow\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\ge-6\Leftrightarrow\frac{48\sqrt{x}-4}{7\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}0\le\sqrt{x}\le\frac{1}{12}\\\sqrt{x}>\frac{1}{7}\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x\le\frac{1}{144}\\x>\frac{1}{49}\end{cases}}\)
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 \(0\le x< 4\)
1) \(\frac{1}{\sqrt{2x-1}}\)có nghĩa khi \(\hept{\begin{cases}2x-1\ge0\\\sqrt{2x-1}\ne0\end{cases}}\)
\(\Leftrightarrow2x-1>0\)
\(\Leftrightarrow x>\frac{1}{2}\)
\(\sqrt{5-x}\)có nghĩa khi \(5-x\ge0\Leftrightarrow x\ge5\)
Vậy \(ĐKXĐ:\frac{1}{2}>x\ge5\)
2) \(\sqrt{x-\frac{1}{x}}\)có nghĩa khi \(\hept{\begin{cases}x-\frac{1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x^2}{x}-\frac{1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{x^2-1}{x}\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2-1\ge0\\x>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2\ge1\\x>0\end{cases}}\)
Vậy \(ĐKXĐ:x\ge1\)
3) \(\sqrt{2x-1}\)có nghĩa khi \(2x-1\ge0\) \(\Leftrightarrow x\ge\frac{1}{2}\)
\(\sqrt{4-x^2}\)có nghĩa khi \(4-x^2\ge0\Leftrightarrow x^2\le4\Leftrightarrow x\le2\)
Vậy \(ĐKXĐ:\frac{1}{2}\le x\le2\)
4) \(\sqrt{x^2-1}\)có nghĩa khi \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow x\ge1\)
\(\sqrt{9-x^2}\)có nghĩa khi \(9-x^2\ge0\Leftrightarrow x^2\le9\Leftrightarrow x\le3\)
Vậy \(ĐKXĐ:1\le x\le3\)
ĐKXĐ : \(x\ge1;x\ne2;x\ne3\)
\(P=\left[\frac{\sqrt{x}+\sqrt{x-1}}{1}-\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}\right].\frac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\left(\sqrt{x}-\sqrt{2}\right)}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
\(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\)
\(P=\frac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\)