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\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{3}=\dfrac{-x-\sqrt{x}-1+x+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{3}{x+\sqrt{x}+1}\text{≤}\dfrac{3}{1}=3\) ( x ≥ 0 ; x # 1 )
⇒ \(A_{Max}=3."="\) ⇔ \(x=0\left(TM\right)\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Câu 1:
a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
=>a>4
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
a) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
b) Khi x=5+2\(\sqrt{3}\Leftrightarrow P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{16-12}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{2\left(2-\sqrt{3}\right)}{4}=\dfrac{2-\sqrt{3}}{2}\)
c) Ta có \(\left|A\right|\le1\Leftrightarrow\left|\dfrac{1}{x-1}\right|\le1\Leftrightarrow\dfrac{1}{\left|x-1\right|}\le1\Leftrightarrow\left|x-1\right|\ge1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1\ge1\\1-x\le1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
Kết hợp với ĐK
Vậy x\(\le0\) hoặc \(x\ge2\) thì \(\left|A\right|\le1\)
a) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
b) Khi x=5+2\(\sqrt{3}\Leftrightarrow P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{16-12}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{2\left(2-\sqrt{3}\right)}{4}=\dfrac{2-\sqrt{3}}{2}\)
c) Ta có \(\left|A\right|\le1\Leftrightarrow\left|\dfrac{1}{x-1}\right|\le1\Leftrightarrow\dfrac{1}{\left|x-1\right|}\le1\Leftrightarrow\left|x-1\right|\ge1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1\ge1\\1-x\le1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
Kết hợp với ĐK
Vậy x\(\le0\) hoặc \(x\ge2\) thì \(\left|A\right|\le1\)
