Question

Cho Y = \(\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)\)

a. rút gọn

b. Tìm x để Y < \(\dfrac{1}{2}\)

( giúp mk vs ạ , mk đang cn gấp)

Answer 1

a) ĐKXĐ: \(x>0;x\ne1;x\ne4\), rút gọn: \(Y=\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)+1}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-1\right)^2-1}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-3\sqrt{x}+3+1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{x-2\sqrt{x}+1-1}=\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Answer 2

\(a.Y=\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)=\dfrac{x-4\sqrt{x}+3+1}{\sqrt{x}-1}:\dfrac{x-2\sqrt{x}+1-1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\left(x>0;x\ne1;x\ne4\right)\)

\(b.Y< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-4}{2\sqrt{x}}< 0\)

Do : \(2\sqrt{x}>0\)\(\Leftrightarrow\sqrt{x}-4< 0\Leftrightarrow x< 16\)

Kết hợp với ĐKXĐ : \(0< x< 16\left(x\ne1;x\ne4\right)\)

Answer 3

b.\(Y< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{1}{2}< 0\Leftrightarrow\dfrac{2\sqrt{x}-4-\sqrt{x}}{2\sqrt{x}}< 0\Leftrightarrow\dfrac{\sqrt{x}-4}{2\sqrt{x}}< 0.Vìx>0\Rightarrow\sqrt{x}-4< 0\Leftrightarrow\sqrt{x}< 4\Leftrightarrow x< 16\)Vậy với mọi x sao cho 0<x<16(x\(\notin\left\{1;4\right\}\)) thì Y<\(\dfrac{1}{2}\)

Answer 4

a) ĐK: \(x\ge0\)

\(Y=\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)\)

\(Y=\left(\dfrac{x-4\sqrt{x}+3+1}{\sqrt{x}-1}\right):\left(\dfrac{x-2\sqrt{x}+1-1}{\sqrt{x}-1}\right)\)

\(Y=\left(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-1}\right):\left(\dfrac{x-2\sqrt{x}}{\sqrt{x}-1}\right)\)

\(Y=\left(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1}\right)\)

\(Y=\left(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(Y=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b) Để Y<\(\dfrac{1}{2}\)

Thì \(Y=\dfrac{\sqrt{x}-2}{\sqrt{x}}< \dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}-4< \sqrt{x}\)

\(\Leftrightarrow\sqrt{x}< 4\Rightarrow x< 16\)