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Mình nghĩ nên bổ sung x nguyên
\(A=\frac{3\sqrt{x}-2}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\frac{5}{\sqrt{x}+1}\)
Để A nguyên thì \(\frac{5}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\in\left\{1;5\right\}\Leftrightarrow x=0\)
Thay x = 0 vào thì ta có A âm vậy ............
a) Ta có: \(x=9\)thỏa mãn đk
\(\Rightarrow\)Thay \(x=9\)vào biểu thức ta được:
\(A=\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=\frac{-9}{2}\)
b) Với x thỏa mãn ĐKXĐ thì ta có:
\(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{1}{\sqrt{x}+2}+\frac{x+14}{x-4}-\frac{4}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)+\left(x+12\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2+x+12-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
1. x = 9 => A = \(\frac{3\sqrt{9}}{1-\sqrt{9}}=\frac{9}{-2}=-\frac{9}{2}\)
2. \(B=\frac{1}{\sqrt{x}+2}-\frac{x+12}{4-x}-\frac{4}{\sqrt{x}-2}=\frac{\sqrt{x}-2+x+12-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
3. \(AB>-\frac{3}{4}\) <=> \(\frac{3\sqrt{x}}{1-\sqrt{x}}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+2}>-\frac{3}{4}\)
<=> \(-\frac{3\sqrt{x}}{\sqrt{x}+2}+\frac{3}{4}>0\)
<=> \(\frac{12\sqrt{x}-3\sqrt{x}-4}{4\left(\sqrt{x}+2\right)}< 0\)
<=> \(\frac{9\sqrt{x}-4}{4\sqrt{x}+8}< 0\)
Do \(4\sqrt{x}+8>0\)với mọi x => \(9\sqrt{x}-4< 0\) <=> \(x< \frac{16}{81}\)
\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right)\div\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để A > 0
=> \(\frac{\sqrt{x}-1}{\sqrt{x}}>0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>0\end{cases}}\Leftrightarrow x>1\)
2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}< 0\end{cases}}\)( dễ thấy trường hợp này không xảy ra :> )
Vậy với x > 1 thì A > 0
\(a,Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}};x>0;x\ne1;x\ne4\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
\(a,\)\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{x-1}\)\(\left(đpcm\right)\)
\(b,Q=\frac{2}{x-1}\)
\(Q\in Z\Leftrightarrow\frac{2}{x-1}\in Z\Rightarrow x-1\inƯ_2\)
Mà \(Ư_2=\left\{\pm1;\pm2\right\}\)
TH1 : \(x-1=-1\Rightarrow x=0\)
TH2 : \(x-1=1\Rightarrow x=2\)
TH3 : \(x-1=-2\Rightarrow x=-1\)
TH4 :\(x-1=2\Rightarrow x=3\)
\(\Rightarrow\)x nguyên lớn nhất là 3 để Q là số nguyên
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
a)
Ta có:\(\left|P\right|>P\Leftrightarrow P< 0\) tức là \(\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
b)
\(\left|P\right|=P\Leftrightarrow P\ge0\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)
Ta luôn có 1 tính chất rất quan trọng sau: \(|P|\ge P\forall P\inℝ\) (*)
DẤU "=" XẢY RA <=> \(P\ge0\)
** Mà để \(|P|>P\Rightarrow P< 0\) (KO CÓ DẤU "=" XẢY RA)
=> \(\frac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
Mà do \(x>0\Rightarrow3\sqrt{x}>0\)
=> \(\sqrt{x}-2< 0\)
=> \(0< x< 4\) VÀ \(x\ne1\) (ĐKXĐ)
** ĐỂ \(|P|=P\)
=> DẤU "=" CỦA BĐT (*) PHẢI XẢY RA
<=> \(P\ge0\) DO: \(x>0\Rightarrow P>0\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-2}>0\)
MÀ: \(\sqrt{x}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy x > 4 thì \(|P|=P\)