Đáp án:
`x=(7±3\sqrt{5})/2`
Giải thích các bước giải:
` x-4\sqrt{x}+1/x-4/\sqrt{x}+5=0`
`->x+1/x-4(\sqrt{x}+1/\sqrt{x})+5=0`
Đặt `a=\sqrt{x}+1/\sqrt{x}`
`->a^2=x+1/x+2`
`->x+1/2=a^2-2`
`->` Phương trình trở thành
`a^2-2-4a+5=0`
`->a^2-4a+3=0`
`->a^2-a-3a+3=0`
`->a(a-1)-3(a-1)=0`
`->(a-3)(a-1)=0`
`->`\(\left[ \begin{array}{l}a-3=0\\a-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}a=3\\a=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\sqrt{x}+\dfrac{1}{\sqrt{x}}=3\\\sqrt{x}+\dfrac{1}{\sqrt{x}}=1\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x+1=3\sqrt{x}\\x+1=\sqrt{x}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x-3\sqrt{x}+\dfrac{9}{4}-\dfrac{5}{4}=0\\x-\sqrt{x}+\dfrac{1}{2}+\dfrac{1}{2}=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}(\sqrt{x}-\dfrac{3}{2})^2=\dfrac{5}{4}\\(\sqrt{x}-\dfrac{1}{2})^2+\dfrac{1}{2}=0 \text{ (Vô lý)}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\sqrt{x}-\dfrac{3}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}-\dfrac{3}{2}=\dfrac{-\sqrt{5}}{2}\end{array} \right.\)
`->\sqrt{x}=(3±\sqrt{5})/2`
`->x=(7±3\sqrt{5})/2(TM)`
