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\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x+\sqrt{x}-2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x+\sqrt{x}\)
a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)
\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)(1)
b) Ta có: \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào biểu thức (1), ta được:
\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)
\(Đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\left(\sqrt{x}-1\right)^2>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>1\end{cases}\Rightarrow}x>1}\)
\(C=\)\(\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{x\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{3\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{x\sqrt{x}\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{2x.\sqrt{x}}{x\sqrt{x}\left(\sqrt{x-1}\right)^2}\)
\(=x\left(\sqrt{x}-1\right)^2+3\left(\sqrt{x}-1\right)^2+x\sqrt{x}\left(\sqrt{x}-1\right)^2+2x.\sqrt{x}\)
.....
a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(=\dfrac{2\left(x+\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2\sqrt{x}+2+2\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{3x-8\sqrt{x}+5+2x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
\(\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\left(x\ne1,x>=0\right)\\ =\dfrac{2}{\sqrt{x}-1}+\dfrac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\dfrac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{2x+2\sqrt{x}+2+2\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5x-5\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)