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a)Đkxđ : x#1 , x > 0
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Q=\(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}X\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
Q=\(\dfrac{x-1}{\sqrt{x}}\)
b)Thay x = 2\(\sqrt{2}\)+3 vào phương trình ta được :
Q=\(\dfrac{2\sqrt{2}+3-1}{\sqrt{2\sqrt{2}+3}}\)
Q=\(\dfrac{2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}+1\right)}^2}\)
Q=\(\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
Q= 2
a: \(P=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{4\sqrt{x}}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}+4\sqrt{x}\left(x+1\right)}{\left(x^2-1\right)}\right)-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x\left(x+1\right)}{x^2-1}-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x^2+4x-5x}{x^2-1}\)
\(=\dfrac{4x^2}{x^2-1}\)
Khi x=4 thì \(P=\dfrac{4\cdot16}{16-1}=\dfrac{64}{15}\)
b: Để P/Q=0 thì P=0
=>x=0
\(\text{a) }\dfrac{1}{\sqrt{x-1}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-1\ge0\\\sqrt{x-1}\ne0\end{matrix}\right.\\ \Rightarrow x-1>0\\ \Rightarrow x>1\)
\(\text{b) }\dfrac{1}{\sqrt{x-\sqrt{2x-1}}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}x-\sqrt{2x-1}\ge0\\\sqrt{x-\sqrt{2x-1}}\ne0\end{matrix}\right.\\ \Rightarrow x-\sqrt{2x-1}>0\\ \Rightarrow x>\sqrt{2x-1}\\ \Rightarrow x^2>2x-1\\ \Rightarrow x^2-2x+1>0\\ \Rightarrow\left(x-1\right)^2>0\\ \Rightarrow\left|x-1\right|>0\\ \Rightarrow\left[{}\begin{matrix}x-1< 0\\x-1>0\end{matrix}\right.\\ \Rightarrow x-1\ne0\\ \Rightarrow x\ne1\)
\(c\text{) }\sqrt{-\dfrac{1}{x}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{x}\ge0\\x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}< 0\\x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< 0\left(\text{Vì }1>0\right)\\x\ne0\end{matrix}\right.\Rightarrow x< 0\)
\(\text{d) }\sqrt{\dfrac{a+1}{a^2}}\\ \text{Để biểu thức có nghĩa }\\ thì\Rightarrow\left\{{}\begin{matrix}\dfrac{a+1}{a^2}\ge0\\a^2\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+1\ge0\left(\text{Vì }a^2>0\right)\\a\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ge-1\\a\ne0\end{matrix}\right.\)
a. \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{x-1}{-\left(x+4\right)}=\dfrac{4\sqrt{x}}{x+4}\)
b. \(\:B=\dfrac{4\sqrt{3+2\sqrt{2}}}{3+2\sqrt{2}+4}=\dfrac{4+4\sqrt{2}}{7+2\sqrt{2}}=\dfrac{\left(4+4\sqrt{2}\right).\left(7-2\sqrt{2}\right)}{\left(7+2\sqrt{2}\right).\left(7-2\sqrt{2}\right)}=\dfrac{12+20\sqrt{2}}{41}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\x-1\ne0\\\sqrt{x}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge0\end{matrix}\right.\)
b) \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}+1}-\dfrac{2}{x-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)}\)c)\(B=A\left(x-1\right)=\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)}\left(x-1\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)(Vì \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\))
=> MinB =\(-\dfrac{1}{4}\) khi x= \(\dfrac{1}{4}\)
ĐKXĐ : \(x\ge0\) và \(x\ne\dfrac{1}{9}\)
\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{5\sqrt{x}\left(\sqrt{x}+1\right)}{5\left(3\sqrt{x}-1\right)}=\dfrac{6\left(3\sqrt{x}-1\right)}{5\left(3\sqrt{x}-1\right)}\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow5x+5\sqrt{x}-18\sqrt{x}+6=0\)
\(\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow5x-10\sqrt{x}-3\sqrt{x}+6=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(5\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\5\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{25}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{25};4\right\}\)
Học tốt !
\(\dfrac{\sqrt{x-1}}{x^2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ge0\\x^2\ne0\end{matrix}\right.\Leftrightarrow x\ge1\)
\(\sqrt{\dfrac{x}{\left(x-1\right)^2}}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-1\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ge0\)
\(\sqrt{x+5}-\sqrt{2x+1}\)
ĐKXĐ:\(\left\{{}\begin{matrix}x+5\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Leftrightarrow x\ge\dfrac{-1}{2}\)
\(\sqrt{3-x^2}\)
ĐKXĐ: \(3-x^2\ge0\Leftrightarrow x\le\pm\sqrt{3}\)
hình như đề là tìm x thỏa mãn: \(\dfrac{1}{Q}+P\ge4\) thì phải
nếu vậy thì ta có \(\dfrac{1}{Q}+P\) = \(\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\)
= \(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
≥ 2+2 = 4(bđt Cauchy)
vậy \(\dfrac{1}{Q}+P\ge4\)