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a: \(Q=\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(=\dfrac{a+2\sqrt{a}+1}{a-\sqrt{a}}\)
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
a, A= \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\left(\sqrt{x}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)}+\frac{x}{\sqrt{x}+2}\right)\)
A=\(\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\frac{\sqrt{x}+x}{\left(\sqrt{x}+2\right)}\right)\)
A=\(\frac{1}{x+2\sqrt{x}}\)
b, A >= \(\frac{1}{3\sqrt{x}}\)
=> \(\frac{1}{x+2\sqrt{x}}\) >= \(\frac{1}{3\sqrt{x}}\)
=> x <= -1 , x >= 4 (x khác 0)
\(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}\) đk: \(x\ge0,x\ne1\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}:\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right]-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{\left(x+1\right)-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}.\dfrac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)^2}-\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b)Để P<4 \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}< 4\) \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-4< 0\) \(\Leftrightarrow\dfrac{\sqrt{x}+2-4\left(\sqrt{x}-1\right)}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\dfrac{6-3\sqrt{x}}{\sqrt{x}-1}< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-3\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}6-3\sqrt{x}< 0\\\sqrt{x}-1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 1\\\sqrt{x}>2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0\le x< 1\\x>4\end{matrix}\right.\)
Vậy...
c)\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\) \(=1+\dfrac{3}{\sqrt{x}-1}\)
Để P nguyên khi \(\dfrac{3}{\sqrt{x}-1}\) nguyên
\(x\in Z\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\in I\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-1\in Z\\\sqrt{x}-1\in I\end{matrix}\right.\)
Tại \(\sqrt{x}-1\in I\Rightarrow\dfrac{3}{\sqrt{x}-1}\notin Z\) (L)
Tại\(\sqrt{x}-1\in Z\) .Để \(\dfrac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;-2;4\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4\right\}\) \(\Leftrightarrow x\in\left\{0;4;16\right\}\) (tm)
ĐKXĐ : x > 0 , x khác 1
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\div\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\div\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\div\left[\dfrac{x+\sqrt{x}-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)
\(=2\div\dfrac{1}{\sqrt{x}-1}=2\sqrt{x}-2\)
b) Dễ thấy ∀ x ≥ 0 thì \(2\sqrt{x}-2\) nguyên
Kết hợp với ĐKXĐ => Với \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)thì A đạt giá trị nguyên
b, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
Ta có : \(B=\dfrac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4-x+2\sqrt{x}-4+x+2}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+2}{\sqrt{x}}\)
b) Ta có: \(B=\dfrac{x\sqrt{x}-8}{x-2\sqrt{x}}-\dfrac{x\sqrt{x}+8}{x+2\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+4}{\sqrt{x}}-\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}+\dfrac{x+2}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+x+2}{\sqrt{x}}\)
c) Ta có: \(C=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3-5+\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)