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\(P=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
\(A=\left(\dfrac{\sqrt{x}}{x-\sqrt{x}}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)\left(x>0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{x-\sqrt{x}+1}{x+1}\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{x+1}{x-\sqrt{x}+1}\)
\(=\dfrac{x+1-2}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(A=\left[\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}-\frac{2}{(\sqrt{x}-1)(x+1)}\right]:\frac{x-\sqrt{x}+1}{x+1}\)
\(=\left[\frac{1}{\sqrt{x}-1}-\frac{2}{(\sqrt{x}-1)(x+1)}\right].\frac{x+1}{x-\sqrt{x}+1}=\frac{x+1-2}{(\sqrt{x}-1)(x+1)}.\frac{x+1}{x-\sqrt{x}+1}=\frac{x-1}{(\sqrt{x}-1)(x-\sqrt{x}+1)}=\frac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)
\(=-\dfrac{x}{5-\sqrt{x}}\)
\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)
a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow B=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(\Leftrightarrow B=\dfrac{2-\sqrt{x}}{3\sqrt{x}}\)
b) \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\) (*)
Thay (*) vào B , ta được : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)
1. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\left(x>0\right)\)
\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
2. Để \(P>0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\sqrt{2}\\x>\sqrt{-1}\left(L\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< \sqrt{2}\\x< \sqrt{-1}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)
Vậy \(P>0\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-2}{\sqrt{x}}\)
\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)
\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)
b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\) mới đúng bn ơi
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