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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 ............
\(a,Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}};x>0;x\ne1;x\ne4\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
\(a,\)\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\)\(\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{x-1}\)\(\left(đpcm\right)\)
\(b,Q=\frac{2}{x-1}\)
\(Q\in Z\Leftrightarrow\frac{2}{x-1}\in Z\Rightarrow x-1\inƯ_2\)
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) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-x}-\frac{\sqrt{x}+2}{1-x}\)
\(\Leftrightarrow M=\frac{1}{1-\sqrt{x}}-\frac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow M=\frac{1+\sqrt{x}-\sqrt{x}-2}{1-x}\)
\(\Leftrightarrow M=\frac{-1}{1-x}\)
\(\Leftrightarrow M=\frac{1}{x-1}\)
b) Để M nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-1}\inℤ\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Mà \(x>0\)
Vậy để M nguyên \(\Leftrightarrow x=2\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
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 \(|P|=P\)