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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
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\(R=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{3x-5\sqrt{x}}{4-x}\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}-1\right)\left(ĐK:x\ge0,x\ne4\right)\\ =\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{3x-5\sqrt{x}}{\sqrt{x}^2-2^2}\right):\dfrac{2\sqrt{x}-1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+3x-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}-2}{2\sqrt{x}-1-\sqrt{x}+2}\\ =\dfrac{3x-6\sqrt{x}+x+2\sqrt{x}+3x-5\sqrt{x}}{\sqrt{x}+2}.\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{7x-9\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
Bạn xem lại đề nhé, rút gọn thường ra kết quả rất đẹp chứ không dài như kết quả này đâu ạ.
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1: Thay \(x=\dfrac{1}{25}\) vào C, ta được:
\(C=\left(\dfrac{1}{5}+2\right):\left(\dfrac{1}{5}-3\right)=\dfrac{11}{5}:\dfrac{-14}{5}=-\dfrac{11}{14}\)
2: Để C=-2 thì \(\sqrt{x}+2=-2\left(\sqrt{x}-3\right)\)
\(\Leftrightarrow\sqrt{x}+2+2\sqrt{x}-6=0\)
\(\Leftrightarrow3\sqrt{x}=4\)
hay \(x=\dfrac{16}{9}\)
Để \(C=\dfrac{7}{5}\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{7}{5}\)
\(\Leftrightarrow7\sqrt{x}-21=2\sqrt{x}+10\)
\(\Leftrightarrow5\sqrt{x}=31\)
hay \(x=\dfrac{961}{25}\)
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biểu thức e viết liền quá khó phân biệt ví dụ như x +1 -\(\frac{2\sqrt{x}}{\sqrt{x-1}}\)hay là x +\(\frac{1-\sqrt{2x}}{\sqrt{x-1}}\)
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a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=9/4 thì A=3/2:1/2=3/2*2=3
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a,Để \(\sqrt{x^2-8x-9}\) có nghĩ thì
\(x^2-8x-9\ge0\)
\(\Leftrightarrow x^2+x-9x-9\ge0\)
\(\Leftrightarrow x\left(x+1\right)-9\left(x+1\right)\ge0\)
\(\Leftrightarrow\left(x+1\right)\left(x-9\right)\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1\ge0\\x-9\ge0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\ge-1\\x\ge9\end{cases}\Rightarrow}x\ge9\)
\(or\orbr{\begin{cases}x+1\le0\\x-9\le0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\le-1\\x\le9\end{cases}\Rightarrow}x\le-1\)
\(Để\sqrt{4-9x^2}\text{có nghĩa}\)
\(\Rightarrow4-9x^2\ge0\)
\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)\ge0\)
\(\Leftrightarrow-\frac{2}{3}\le x\le\frac{2}{3}\)
a) \(\frac{1}{\sqrt{x^2-8x+15}}\)DK : \(x^2-8x+15>0\Rightarrow x< 3\)hoặc \(x>5\)
b) \(\sqrt{2}-\sqrt{x-1}\)DK : \(x-1\ge0\Rightarrow x\ge1\)