cho biểu thức A=\(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
a, nêu điều kiện xác định và rút gọn biểu thức A
b, tìm giá trị của x đẻ A=\(\dfrac{1}{3}\)
c, tìm giá trị lớn nhất của biểu thức P=A-\(9\sqrt{x}\)
các bạn giúp mình vs quan trọng là câu c
Câu a : ĐK : \(x>1\)
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu b : \(A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Leftrightarrow2\sqrt{x}=3\Leftrightarrow x=\dfrac{9}{4}\)
Câu c : \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(9\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\le1-2\sqrt{9}=-5\)
Vậy GTLN là -5 . Khi \(x=\dfrac{1}{9}\)
$a) ĐK:$ \(x>1\)
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right].\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
$b)$ \(A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Leftrightarrow2\sqrt{x}=3\Leftrightarrow x=\dfrac{9}{4}\)
$c)$ \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(9\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\le1-2\sqrt{9}=-5\)
Vậy $GTLN$ là $-5$ . Khi \(x=\dfrac{1}{9}\)
$a) ĐK: \(x\ne1,x\ne0\)
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right].\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\\ A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
$b)$ \(A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\Leftrightarrow2\sqrt{x}=3\Leftrightarrow x=\dfrac{9}{4}\)
$c)$ \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(9\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)\le1-2\sqrt{9}=-5\)
Vậy $GTLN$ là $-5$ . Khi \(x=\dfrac{1}{9}\)
