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a)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b)
\(Q< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}< 0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\ \Leftrightarrow0< x< 4\)
a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)
c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
a) P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\) với x > 0 và x≠4
=\(\left(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{x-4}\right).\dfrac{x-4}{2\sqrt{x}}\)
=\(\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\).\(\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{x}{\sqrt{x}}\)
b) \(\dfrac{x}{\sqrt{x}}\) >3
<=> x> \(3\sqrt{x}\)
<=> x>9
a: \(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b: Để P>3 thì \(\sqrt{x}>3\)
hay x>9
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
\(B=\left(\dfrac{x}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\\ ĐKXĐ:x\ge0;x\ne4\\ =\left(\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{10-x}{\sqrt{x}+2}\right)\\ =\dfrac{x-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{x-4+10-x}{\sqrt{x}+2}\\ =\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{6}{\sqrt{x}+2}\\ =\dfrac{x-3\sqrt{x}+2\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{6}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}\)
b) Với \(x\ge0;x\ne4\)
\(\Rightarrow\) Để B>0
thì \(\Rightarrow\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{6\left(\sqrt{x}-2\right)}>0\)
\(\Rightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}>0\left(Vì\text{ }\sqrt{x}+2;6>0\right)\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\\sqrt{x}-2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-2< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}>3\\\sqrt{x}< 2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge9\\0< x< 4\end{matrix}\right.\)
b) \(B>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0,do-1< 0\Rightarrow B>0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
a) Ta có
\(P=\left(\dfrac{x}{x-2\sqrt{x}}+\dfrac{x}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\) (ĐKXĐ : \(x\ne\pm4;x\ne0\) )
\(P=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2}\)
\(P=\dfrac{x+x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\sqrt{x}}=\sqrt{x}-2\)
b) \(P>0\)
\(\Rightarrow\sqrt{x}-2>0\)
\(\Rightarrow x>4\)
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)
thank