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\(a.P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{10\sqrt{x}-2x}\left(x>0,x\ne4,x\ne25\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right].\dfrac{x-4}{10\sqrt{x}-2x}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{10\sqrt{x}-2x}\)
\(=\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}\left(5-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}}{5-\sqrt{x}}\)
\(b.\) Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(\dfrac{\sqrt{\dfrac{1}{4}}}{5-\sqrt{\dfrac{1}{4}}}=\dfrac{0,5}{5-0,5}=\dfrac{1}{9}\)
Vậy ......................
\(c.P< -1\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{5-\sqrt{x}}< -1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+5-\sqrt{x}}{5-\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{5}{5-\sqrt{x}}< 0\)
\(\Leftrightarrow5-\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}>5\)
\(\Leftrightarrow x>25\left(tm\right)\)
Vậy ...................
\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)^2}\)
\(P=-\dfrac{1}{3}\)
\(\Rightarrow\left(\sqrt{x}+3\right)^2=3\sqrt{x}+3\)
\(\Leftrightarrow x-\sqrt{x}+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow x=9\left(Vì\sqrt{x}+2>0\right)\)
\(P=-\left(\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}+3\right)^2}\right)=-\left(\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)^2}\right)< -3< -1\)
Điều kiện: x>2
P= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{2}+2}{\sqrt{x}-1}\right)\)
P= \(\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
P= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
P= \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) P= \(\dfrac{1}{4}\)
⇔\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}\) =\(\dfrac{1}{4}\)
⇔\(4\sqrt{x}-8=3\sqrt{x}\)
⇔\(\sqrt{x}=8\)
⇔x=64 (TM)
Vậy X=64(TMĐK) thì P=\(\dfrac{1}{4}\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}-1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}+1}-1\right):\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}+1\right)+\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}-1\right)-2x+1}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}:\left(\dfrac{2x-1+\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)-\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}\right)\)
\(=\dfrac{x\sqrt{2}+\sqrt{x}+\sqrt{2x}+1+2x-\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{2x-1}:\dfrac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{x}+2}{-2-2\sqrt{x}}\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\left(ĐKXĐ:x\ge0;x\ne9\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}-3}{x-9}\)
\(b,M=P:Q\)
\(=\dfrac{-3\sqrt{x}-3}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
Ta thấy: \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}+3\ge3\forall x\)
\(\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\forall x\)
\(\Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}=-1\)
hay \(M\ge-1\)
#Toru
a) ĐKXĐ: \(x>0,x\ne1\)
\(P=\dfrac{x-2\sqrt{x}}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\dfrac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{1+2x-2\sqrt{x}}{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}\)
\(=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\dfrac{1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-2\sqrt{x}\right)\sqrt{x}+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1+2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}+x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(x+\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
b) Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+2>0\\x+\sqrt{x}+1>0\end{matrix}\right.\Rightarrow P>0\)
Vì \(x>0\Rightarrow2x+\sqrt{x}>0\Rightarrow2x+2\sqrt{x}+2-\left(\sqrt{x}+2\right)>0\)
\(\Rightarrow2\left(x+\sqrt{x}+1\right)>\sqrt{x}+2\Rightarrow\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}< 2\)
mà P nguyên \(\Rightarrow\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}=1\Rightarrow\sqrt{x}+2=x+\sqrt{x}+1\)
\(\Rightarrow x-1=0\Rightarrow x=1\) mà \(x\ne1\Rightarrow\) không có x để P nguyên
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)
a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=-\dfrac{1}{2\sqrt{x}-3}\)
b: \(x=\dfrac{2}{2-\sqrt{3}}=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)
Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)
c: C=-1
=>\(2\sqrt{x}-3=1\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
d: C>0
=>\(2\sqrt{x}-3< 0\)
=>\(\sqrt{x}< \dfrac{3}{2}\)
=>\(0< =x< \dfrac{9}{4}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)
Sửa đề: