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a: ĐKXĐ: x>0; x<>1
b: \(A=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
c: A nguyên
=>x-1 thuộc {1;-1;2;-2}
=>x thuộc {2;3}
ĐKXĐ: \(x>0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b.
\(P=x-\sqrt{x}+1=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
a) đk: \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\sqrt{x}-1>0\\x>0\end{matrix}\right.=>\sqrt{x}>\pm1\)
rút gọn pt
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) \(\dfrac{\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2x+\sqrt{x}\right)\left(\sqrt{x}-1\right)\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(x-1\right)x\left(x+1\right)}{x\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\)
\(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}\) 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)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}-2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{2a}\)
\(=\sqrt{a}+2\)
b: A-2<0
=>\(\sqrt{a}+2-2< 0\)
=>\(\sqrt{a}< 0\)
=>\(a\in\varnothing\)
c: Bạn ghi đầy đủ đề đi bạn
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
Lời giải:
ĐKXĐ: $x>0; x\neq 4$
Sửa lại đề 1 chút.
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{2}{\sqrt{x}+2}\)
\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)
Với mọi $x>0$ thì hiển nhiên $B>0$. Mặt khác, $\sqrt{x}+2\geq 2$ nên $B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}=\frac{7}{3}$
Vậy $0< B\leq \frac{7}{3}$. $B$ đạt giá trị nguyên thì $B=1;2$
$B=1\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=1$
$\Leftrightarrow x=\frac{64}{9}$ (thỏa mãn)
$B=2\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=2$
$\Leftrightarrow x=\frac{1}{9}$ (thỏa mãn)
a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1
A=(\(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
A=\(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
A=\(\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)=\(\dfrac{2}{a-1}\)
b) Để A\(\in\)Z\(\Rightarrow\)x-1\(\in\) Ư(2)=\(\left\{-1,1,-2,2\right\}\)
vì x\(\ge\)0,x\(\ne\)1 nên x\(\in\)\(\left\{-1,0,2,3\right\}\)