rút gọn P
P = \(\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+2\sqrt{x}-1}\)
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13 tháng 7 2016
\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{x+\sqrt{x}}-\frac{2}{1-x}\right)\) (ĐKXĐ : \(x>0;x\ne1;x\ne\frac{1}{9}\) )
\(=\left[\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\frac{\sqrt{x}-1}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{3\sqrt{x}-1}\)
3 tháng 9 2016
\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)
\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\frac{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
10 tháng 8 2015
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+9\right)}\right).\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}\)
\(=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\sqrt{x}-2}=\frac{-3\sqrt{x}-3}{2x-8\sqrt{x}+6}\)
Nếu đề ko sai thì đấy là kết quả
BP
29 tháng 10 2016
ĐKXĐ:\(\sqrt{x}\ge0\Leftrightarrow x\ge0\)
Rút gọn: P=\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}=\frac{2\sqrt{x}\left(x-1\right)^2}{2\left(x-1\right)\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)=x-1\)
NA
5 tháng 10 2018
\(\left(x-1\right)-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(=\left(x-1\right)-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=x-1-2\sqrt{x}+\sqrt{x}+1\)
\(=x-\sqrt{x}\)
P=\(\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+2\sqrt{x}-1}\)(đk :\(x\ge0,x\ne\frac{1}{9},x\ne1\))
=\(\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+3\sqrt{x}-\sqrt{x}-1}=\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\)(1)
TH1 : \(0\le\sqrt{x}\le1\)
Từ (1)=> \(P=\frac{2\sqrt{x}+1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}=\frac{1}{3\sqrt{x}-1}\)
TH2: x>1
Từ (1) => \(P=\frac{2\sqrt{x}+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}+1}\)
Vậy với \(0\le x\le1\) => \(P=\frac{1}{3\sqrt{x}-1}\)
x>1=> P=\(\frac{1}{\sqrt{x}+1}\)
\(P=\frac{2\sqrt{x}+\left|\sqrt{x}-1\right|}{3x+2\sqrt{x}-1}\)
ĐK: \(x\ge0;x\ne\frac{1}{9}\)
\(TH_1:\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)
\(P=\frac{2\sqrt{x}+\sqrt{x}-1}{3x+2\sqrt{x}-1}\\ =\frac{3\sqrt{x}-1}{3x+3\sqrt{x}-\sqrt{x}-1}\\ =\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}\\ =\frac{1}{1+\sqrt{x}}=\frac{1-\sqrt{x}}{1-x}\)
\(TH_2:\sqrt{x}-1< 0\Leftrightarrow x< 1\)
\(P=\frac{2\sqrt{x}+1-\sqrt{x}}{3x+2\sqrt{x}-1}\\ =\frac{\sqrt{x}+1}{3x+3\sqrt{x}-\sqrt{x}-1}\\ =\frac{\sqrt{x}+1}{\left(3\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}\\ =\frac{1}{3\sqrt{x}-1}\\ =\frac{3\sqrt{x}+1}{9x-1}\)