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\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\) \(ĐKXĐ:x\ne4\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\) \(\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
vậy \(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) \(P=\frac{3.\frac{1}{2}}{\frac{1}{2}+2}=\frac{3}{2}:\frac{5}{2}=\frac{3}{2}.\frac{2}{5}=\frac{3}{5}\)
vậy khi \(x=\frac{1}{4}\)thì \(P=\frac{3}{5}\)
c) \(P< 2\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-2\sqrt{x}-4}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}< 0\)
đến đây làm 4 trường hợp rồi hợp nghiệm là xong
a) A xác định \(\Leftrightarrow\hept{\begin{cases}x^2-2x\ge0\\x-\sqrt{x^2-2x}\ne0\\x+\sqrt{x^2-2x}\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< 0\\x\ge2\end{cases}}\)
b) \(A=\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}=\frac{\left(x^2+x^2-2x+2x\sqrt{x^2-2x}\right)-\left(x^2+x^2-2x-2x\sqrt{x^2-2x}\right)}{x^2-\left(x^2-2x\right)}\)\(=\frac{4x\sqrt{x^2-2x}}{2x}=2\sqrt{x^2-2x}\)
c) \(A< 2\Leftrightarrow2\sqrt{x^2-2x}< 2\Leftrightarrow x^2-2x< 1\Leftrightarrow x^2-2x-1< 0\Leftrightarrow1-\sqrt{2}\le x\le1+\sqrt{2}\)
Kết hợp với điều kiện A xác định được : \(2\le x\le1+\sqrt{2}\)
Vậy \(A< 2\Leftrightarrow2\le x\le1+\sqrt{2}\)
ĐKXĐ: \(x\ge0;\)\(x\ne1\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{x}{\sqrt{x} \left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{x-1}{\sqrt{x}}\)
\(a,ĐKXĐ:x\ne\sqrt{2};-\sqrt{2};x\ne4\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}+\frac{-2-5\sqrt{x}}{x-4}\)
\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)
\(P=\frac{3x-6\sqrt{x}}{x-4}\)
\(b;\)Để P<2
\(\Rightarrow3x-6\sqrt{x}< 2x-8\)
\(\Rightarrow3x-2x< -8+6\sqrt{x}\)
\(\Rightarrow x-6\sqrt{x}< -8\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-6\right)< 8\)
Tìm x là xong
a) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x>4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Ta có : \(P< 2\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-\frac{2\sqrt{x}+4}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}< 0\)
Mà \(\sqrt{x}-4< \sqrt{x}+2\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}+2>0\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{x}< 4\\\sqrt{x}>-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 16\\x>4\end{cases}}\Leftrightarrow4< x< 16\)
Vậy ...