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a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\frac{1}{x-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}-1}\div\frac{2}{x-1}+\frac{1}{\sqrt{x}+1}.\)
=\(\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}-1}\right)\div\frac{2}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}+1}\)
\(=\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{2+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\times\left(\sqrt{x}+1\right)}\)
\(=\frac{1+x}{\sqrt{x}\times\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{\left(1+x\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1+x}{\sqrt{x}}\)
Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn.
Câu 5: B
Câu 3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)
b: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
c: Để P>4 thì \(\sqrt{x}>4\)
=>x>16
\(M=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{x-9}=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2}{\sqrt{x}-3}\)
Để M là số tự nhiên \(\Rightarrow\left\{{}\begin{matrix}2⋮\sqrt{x}-3\\\sqrt{x}-3>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-3\in\left\{2;1;-1;-2\right\}\\x>9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{25;16;4;1\right\}\\x>9\end{matrix}\right.\Rightarrow x\in\left\{25;16\right\}\)
Thế vào M,ta đường \(\left\{{}\begin{matrix}x=25\Rightarrow M=1\\x=16\Rightarrow M=2\end{matrix}\right.\)
\(\Rightarrow M\) có giá trị là số tự nhiên lớn nhất là \(2\) khi \(x=16\)
\(B=\dfrac{\sqrt{x}-1}{3-\sqrt{x}}-\dfrac{9\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1-\sqrt{x}}{\sqrt{x}-3}-\dfrac{9\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)-9\sqrt{x}-5-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1-x-9\sqrt{x}-5-x+3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-2x-6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{-2\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-2\left(\sqrt{x}+2\right)}{\sqrt{x}-3}\)