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a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
1: \(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
2: Để P>=0 thì 3 căn x-1>0
hay x>1/9
Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
\(ĐKXĐ:x\ne1,x\ge0\)
\(Q=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(Q=\left[\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\dfrac{8\sqrt{x}}{9x-1}\right]:\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(Q=\left(\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(Q=\dfrac{3x+2\sqrt{x}}{9x-1}:\dfrac{9\sqrt{x}-3}{9x-1}\)
\(Q=\dfrac{3x+3\sqrt{x}}{9x-1}.\dfrac{9x-1}{9\sqrt{x}-3}\)
\(Q=\dfrac{3x+3\sqrt{x}}{9\sqrt{x}-3}\)
\(Q=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
b. Ta có: \(\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\).
\(\Rightarrow Q=\dfrac{6+2\sqrt{5}+\sqrt{5}+1}{3.\left(\sqrt{5}+1\right)-1}=\dfrac{7+3\sqrt{5}}{3\sqrt{5}+3-1}=\dfrac{7+3\sqrt{5}}{3\sqrt{5}+2}=\dfrac{31+15\sqrt{5}}{41}\)
a) ĐKXĐ: \(x\ge0;x\ne\dfrac{1}{9}\) , rút gọn: \(Q=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\dfrac{3}{3\sqrt{x}+1}=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(3\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)b) Thay x=\(6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) vào Q, ta có: \(Q=\dfrac{6+2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}{3\sqrt{\left(\sqrt{5}+1\right)^2}-1}=\dfrac{6+2\sqrt{5}+\sqrt{5}+1}{3\sqrt{5}+3-1}=\dfrac{3\sqrt{5}+7}{3\sqrt{5}+2}=\dfrac{\left(3\sqrt{5}+7\right)\left(3\sqrt{5}-2\right)}{\left(3\sqrt{5}+2\right)\left(3\sqrt{5}-2\right)}=\dfrac{45-6\sqrt{5}+21\sqrt{5}-14}{45-4}=\dfrac{31+15\sqrt{5}}{41}\)
a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)
\(=1\)
b)ĐK: \(x>0,x\ne9\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
a) Để biểu thức E được xác định thì \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\9x-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)
b) \(E=\left(1-\dfrac{2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{9x-1}\right):\left(\dfrac{9\sqrt{x}+6}{3\sqrt{x}+1}-3\right)=\left[\dfrac{3\sqrt{x}+1-2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\dfrac{9\sqrt{x}+6-9\sqrt{x}-3}{3\sqrt{x}+1}\right)=\left[\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\left(1+\dfrac{1}{3\sqrt{x}-1}\right).\dfrac{3\sqrt{x}+1}{3}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\dfrac{3\sqrt{x}+1}{3}.\dfrac{3\sqrt{x}}{3\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
Bài 1:
a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
b: \(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
a) \(\left(\dfrac{x+2\sqrt{x}-7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)ư
=\(\dfrac{x+2\sqrt{x}-7-\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x +2\sqrt{x}-7-x+\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)
=\(\dfrac{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
b)ta có : \(\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)
để P nguyên thì \(\sqrt{x}-3\inƯ\left(2\right)\Leftrightarrow\sqrt{x}-3\inƯ\left(\pm1,\pm2\right)\)
\(\Rightarrow\sqrt{x}-3=1\Leftrightarrow x=16\left(TM\right)\)
\(\sqrt{x}-3=-1\Leftrightarrow x=4\left(KTM\right)\)
\(\sqrt{x}-3=2\Leftrightarrow x=25\left(TM\right)\)
\(\sqrt{x}-3=-2\Leftrightarrow x=1\left(KTM\right)\)
vậy x\(\in\left\{16,25\right\}\)
= \(\left(\dfrac{9x-1-2\sqrt{x}\left(3\sqrt{x}-1\right)+\sqrt{x}+1}{9x-1}\right):\left(\dfrac{9\sqrt{x}+6-9\sqrt{x}-3}{3\sqrt{x}-1}\right)\)
=\(\dfrac{3x+3\sqrt{x}}{9x-1}.\dfrac{3\sqrt{x}+1}{3}\)
=\(\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)