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ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)
c.
Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có:
\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
a) Ta có: \(P=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
b) Thay \(x=5+2\sqrt{3}\) vào biểu thức \(P=\dfrac{1}{x-1}\), ta được:
\(P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}\)
\(\Leftrightarrow P=\left(\dfrac{1}{\sqrt{3}+1}\right)^2\)
hay \(\sqrt{P}=\dfrac{\sqrt{3}-1}{2}\)
Vậy: Khi \(x=5+2\sqrt{3}\) thì \(\sqrt{P}=\dfrac{\sqrt{3}-1}{2}\)
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 : \(B=\dfrac{2-\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{-\sqrt{3}+1}{3\sqrt{3}+3}\)
Lời giải:
a. \(B=\frac{\sqrt{x}(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{\sqrt{x}+1}{2}=\frac{x-\sqrt{x}-x-\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{\sqrt{x}+1}{2}=\frac{-2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}}{1-\sqrt{x}}\)
b. $B=3\Leftrightarrow \frac{\sqrt{x}}{1-\sqrt{x}}=3$
$\Rightarrow \sqrt{x}=3(1-\sqrt{x})$
$\Leftrightarrow 4\sqrt{x}=3\Leftrightarrow x=\frac{9}{16}$ (tm)
c.
Khi $x=3-2\sqrt{2}=(\sqrt{2}-1)^2\Rightarrow \sqrt{x}=\sqrt{2}-1$
Khi đó:
$B=\frac{\sqrt{x}}{1-\sqrt{x}}=\frac{\sqrt{2}-1}{1-(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-\sqrt{2}}$
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
a, \(P=\left(1-\dfrac{2\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}+x+1}\right)\)(ĐK: \(x\ge0,x\ne-1\))
\(=\left(\dfrac{x-2\sqrt{x}+1}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)+x+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}:\left(\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}+1\)
b, ĐK: \(x\ge0,x\ne-1\)
\(x=2019-2\sqrt{2018}=\left(\sqrt{2018}-1\right)^2\)
Thay \(x=\left(\sqrt{2018}-1\right)^2\)(TMĐK) vào P ta có:
\(P=\sqrt{2018}-1+1=\sqrt{2018}\)
Vậy với \(x=2019-2\sqrt{2018}\) thì \(P=\sqrt{2018}\)