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a) \(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right].\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) \(\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}-\sqrt{x}=3\)
\(\Leftrightarrow2\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{9}{4}\)
a: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)
b: Để A<=3/căn x thì \(\dfrac{x-2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)^2}< =\dfrac{3}{\sqrt{x}}\)
=>\(\dfrac{x-2\sqrt{x}-1-3x+6\sqrt{x}-3}{\left(\sqrt{x}-1\right)^2}< =0\)
=>\(-2x+4\sqrt{x}-4< =0\)
=>\(x-2\sqrt{x}+2>=0\)(luôn đúng)
a: Khi x=9 thì A=(9-2)/(3+2)=7/5
b: \(B=\dfrac{x-\sqrt{x}+2\sqrt{x}+2-4}{x-1}=\dfrac{x+\sqrt{x}-2}{x-1}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
c: P=A*B
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{x-2}{\sqrt{x}+2}=\dfrac{x-2}{\sqrt{x}+1}\)
P=7/4
=>(x-2)/(căn x+1)=7/4
=>4x-8=7căn 7+7
=>4x-7căn x-15=0
=>căn x=3(nhận) hoặc căn x=-5/4(loại)
=>x=9
a) \(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
\(\Leftrightarrow-\sqrt{x}-1=\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)
c) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp đk:
\(\Rightarrow x\in\left\{0\right\}\)
d) \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}< 1\)
\(a,P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,P=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\ \Leftrightarrow2\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\\ c,P\in Z\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\in Z\\ \Leftrightarrow2⋮\sqrt{x}+1\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}+1\in\left\{1;2\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\\ \Leftrightarrow x\in\left\{0;1\right\}\)
\(d,P=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\dfrac{2}{\sqrt{x}+1}>0\left(2>0;\sqrt{x}+1>0\right)\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}< 1\Leftrightarrow P< 1\)
\(e,P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Có \(\sqrt{x}+1\ge1\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\ge1-2=-1\)
\(P_{min}=-1\Leftrightarrow x=0\)
a: \(P=\dfrac{x+\sqrt{x}}{x-\sqrt{x}}\cdot\dfrac{3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}-1}\)
b: Để P=1 thì \(\sqrt{x}-1=3\)
hay x=16
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x+\sqrt{x}}{x\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{3}\)
\(P=\left(\dfrac{x\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}-1\right)}\right).\dfrac{3}{\sqrt{x}+1}\)
\(P=\dfrac{3}{\sqrt{x}-1}\)
\(P=1\)
\(\Leftrightarrow1=\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\left(tm\right)\)
a: Ta có: \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-1}{1}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{x+4}{\sqrt{x}}\)
b: Để A=5 thì \(x+4=5\sqrt{x}\)
\(\Leftrightarrow x=16\)
a. \(A=\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{1-\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{-\sqrt{x}}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x-1+5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)
b. \(A=5\Leftrightarrow\dfrac{x+4}{\sqrt{x}}=5\Leftrightarrow x+4=5\sqrt{x}\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)
Vậy tất cả các x thỏa ycbt là x=1 hoặc x=16
c. \(A>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}>4\Leftrightarrow\dfrac{x+4}{\sqrt{x}}-4>0\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\)
Vì \(\left(\sqrt{x}-2\right)^2\ge0\forall x\) nên \(\left\{{}\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x>0\end{matrix}\right.\)
Vậy tất cả các x thỏa mãn ycbt là x>0 và \(x\ne4\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
\(a,A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(đk:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+2\sqrt{x}+1-3x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2}{\sqrt{x}-1}\)
\(---\)
\(b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với điều kiện của \(x\), ta được:
\(0\le x< 1\)
Vậy: ...
\(Toru\)
a) \(A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2}{\sqrt{x}-1}\)
b) \(A< 0\) khi
\(\dfrac{2}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với đk:
\(0\le x< 1\)