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a: \(P=\dfrac{x+2+x-1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2x+1-x+\sqrt{x}-1}{x\sqrt{x}+1}=\dfrac{x+\sqrt{x}}{x\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
b: |P|=2/3
mà P>=0
nên P=2/3
=>căn x/(x-căn x+1)=2/3
=>2x-2căn x+2=3căn x
=>2x-5*căn x+2=0
=>(căn x-2)(2căn x-1)=0
=>x=4 hoặc x=1/4
a: \(P=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}:\dfrac{3\sqrt{x}+3+\sqrt{x}-1-2}{x-1}\)
\(=\dfrac{-4\sqrt{x}}{3\sqrt{x}}=\dfrac{-4}{3}\)
b: Để \(P=\sqrt{x}-1\) thì \(\sqrt{x}=-\dfrac{1}{3}\)(vô lý)
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b: Ta có: \(P=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
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
a. ĐKXĐ: \(x>0\)
\(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{x+\sqrt{x}}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b. Để \(P=-1\) thÌ \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=-1\)
\(\Leftrightarrow x+\sqrt{x}+1=-\sqrt{x}\)
\(\Leftrightarrow x+2\sqrt{x}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}+1=0\)
\(\Leftrightarrow\sqrt{x}=-1\) ( vô lý )
Vậy không có x thỏa mãn ycbt
c. Ta có \(x=\dfrac{8}{\sqrt{5}-1}-\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8-8\sqrt{5}+8}{5-1}=\dfrac{16}{4}=4\)
Thay x=4 vào P, ta được
\(P=\dfrac{4+\sqrt{4}+1}{\sqrt{4}}=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)
d. \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\) \(\Rightarrow P-3=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-3\)
\(\Rightarrow P-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-1\right)^2\ge0\\\sqrt{x}>0\end{matrix}\right.\) \(\Rightarrow P-3\ge0\Rightarrow P\ge3\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Vậy \(P_{min}=3\) khi \(x=1\)
a) ĐK:\(x\ge0;x\ne9\)
\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b)\(P=-\dfrac{3}{\sqrt{x}+3}\)
Có \(\sqrt{x}+3\ge3;\forall x\ge0\)
\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)
\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)
a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
a: \(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\dfrac{4}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{4\left(\sqrt{x}-2\right)-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\dfrac{-4\sqrt{x}-8}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\left(\sqrt{x}-3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)
b: Q<4
=>Q-4<0
=>\(\dfrac{4\sqrt{x}}{\sqrt{x}-3}-4< 0\)
=>\(\dfrac{4\sqrt{x}-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)
=>\(\dfrac{12}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: 0<x<9 và x<>4
\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)
`b,` Để `Q<4` ta có :
\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) Để \(P=\dfrac{13}{3}\) thì \(3x+3\sqrt{x}+3-13\sqrt{x}=0\)
\(\Leftrightarrow3x-9\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\)