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\(C=\dfrac{\sqrt{\dfrac{4x^2+4x+1}{x}}}{\sqrt{x}\cdot\left|2x^2-x-1\right|}=\dfrac{\left|2x+1\right|}{\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}\cdot\left|\left(x-1\right)\left(2x+1\right)\right|}\)
\(=\dfrac{1}{x\left|x-1\right|}\)
\(\dfrac{2\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\left(ĐK:x\ge0;x\ne4;x\ne9\right)\)
\(=\dfrac{2\sqrt{x}}{x-2\sqrt{x}-3\sqrt{x}+6}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}+7}{x-5\sqrt{x}+6}\)
#Urushi
\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\) (ĐK: \(x\ne9;x\ne4;x\ge0\))
\(=\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9+2x-4\sqrt{x}+\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: x>0; x<>4; x<>1
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x+4\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{4\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{4}\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-3}{x-2\sqrt{x}}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(=\left[\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\dfrac{x-4-\left(x-4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x+4\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}-\sqrt{x}+3}{-4+4\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-3}{4}\)
#Urushi☕
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{x+1}-\dfrac{4-3\sqrt{x}}{x-4\sqrt{x}+4}\right):\left(\dfrac{x-\sqrt{x}}{x\sqrt{x}-2x+\sqrt{x}-2}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-4\sqrt{x}+4\right)+\left(3\sqrt{x}-4\right)\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(x+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+4\sqrt{x}+x-4\sqrt{x}+4+3x\sqrt{x}+3\sqrt{x}-4x-4}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(x+1\right)}{x-\sqrt{x}}\)
\(=\dfrac{4x\sqrt{x}-7x+3\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\cdot\left(4\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-3}{\sqrt{x}-2}\)
Để A>1 thì A-1>0
\(\Leftrightarrow\dfrac{4\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-1}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1\le0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x\le\dfrac{1}{9}\\x>4\end{matrix}\right.\)
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 \(\sqrt{x}-2=0\Rightarrow x=4\)
\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)
\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)
a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)
\(=\dfrac{1}{x}\)
b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)
\(=\dfrac{3x+6}{x-2}\)
\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)
\(=\pm3\sqrt{2}x^2\)
\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)
\(=\pm\left|a\right|\)
\(=4x-2\sqrt{2}+\dfrac{x\sqrt{x+2}}{\sqrt{x+2}}\\ =4x-2\sqrt{2}+x\\ =5x-2\sqrt{2}\)