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a) Ta có:
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(P=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
b) Ta có: \(P>0\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\sqrt{x}}{x}>0\)
\(\Rightarrow\left(x-1\right)\sqrt{x}>0\)
\(\Rightarrow\hept{\begin{cases}x-1>0\\\sqrt{x}>0\end{cases}}\Rightarrow x>1\)
Vậy khi \(x>1\Leftrightarrow P>0\)
c) Ta có: \(P=6\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}=6\)
\(\Leftrightarrow x-1=6\sqrt{x}\)
\(\Leftrightarrow\left(x-1\right)^2=36x\)
\(\Leftrightarrow x^2-38x+1=0\)
\(\Leftrightarrow\left(x^2-38x+361\right)-360=0\)
\(\Leftrightarrow\left(x-19\right)^2-\left(6\sqrt{10}\right)^2=0\)
\(\Leftrightarrow\left(x-19-6\sqrt{10}\right)\left(x-19+6\sqrt{10}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-19-6\sqrt{10}=0\\x-19+6\sqrt{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{cases}}\)
đkxđ là \(x\ne1;x>0\)
\(Q=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
gtnn \(x-\sqrt{x}+1=x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
gtnn 3/4
ý c bạn tự làm nha mk chịu
\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
b.\(Q< 1\)
\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-8< 0\)
\(\Leftrightarrow0\le x< 4\)
Vay de Q<1 thi \(0\le0< 4\)
Bài 1.
\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\frac{x}{x-\sqrt{x}}\)với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
a) \(B=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x}\)
\(B=\frac{4\sqrt{x}\cdot\sqrt{x}}{\left(\sqrt{x}+1\right)x}=\frac{4x}{\left(\sqrt{x}+1\right)x}=\frac{4}{\sqrt{x}+1}\)
b) Để B > 1
=> \(\frac{4}{\sqrt{x}+1}>0\)( với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))
Vì 4 > 0
=> \(\sqrt{x}+1>0\)
<=> \(\sqrt{x}>-1\)( luôn luôn đúng \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)) ( theo ĐKXĐ )
Vậy \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)thì B > 1
Chưa chắc lắm ... Còn câu 2 thì tí nữa mình làm cho
Bài 2.
\(A=2\sqrt{5}-1\)
\(B=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)( x > 0 )
a) \(B=\frac{2}{x-1}\cdot\frac{\sqrt{x^2-2x+1}}{\sqrt{4x^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\sqrt{\left(x-1\right)^2}}{\sqrt{\left(2x\right)^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\left|x-1\right|}{\left|2x\right|}\)
\(B=\frac{2}{x-1}\cdot\frac{x-1}{2x}=\frac{1}{x}\)( vì x > 0 )
b) Để A + B = 0
=> \(\left(2\sqrt{5}-1\right)+\frac{1}{x}=0\)( ĐKXĐ : \(x\ne0\))
<=> \(\frac{1}{x}=-\left(2\sqrt{5}-1\right)\)
<=> \(\frac{1}{x}=1-2\sqrt{5}\)
<=> \(x\times\left(1-2\sqrt{5}\right)=1\)
<=> \(x=\frac{1}{1-2\sqrt{5}}\)( tmđk )
Vậy \(x=\frac{1}{1-2\sqrt{5}}\)