Question

Answer 1

b: \(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

c: \(x-3\sqrt{x}-4=0\)

\(\Leftrightarrow\sqrt{x}-4=0\)

hay x=16

Answer 2

b) \(x-2\sqrt{x}=0\left(đk:x\ge0\right)\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

c) \(x-3\sqrt[]{x}-4=0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)( do \(\sqrt{x}+1\ge1>0\))

Answer 3

b. \(x-2\sqrt{x}=0\)

<=> \(-2\sqrt{x}=-x\)

<=> \(\sqrt{x}=\dfrac{-x}{-2}=\dfrac{x}{2}\)

<=> x = \(\dfrac{x^2}{4}\)

<=> \(\dfrac{4x}{4}=\dfrac{x^2}{4}\)

<=> 4x = x²

<=> x² - 4x = 0

<=> x(x - 4) = 0

<=> \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

c. \(x-3\sqrt{x}-4=0\)

<=> \(-3\sqrt{x}-4=-x\)

<=> \(-3\sqrt{x}=4-x\)

<=> \(\sqrt{x}=\dfrac{4-x}{-3}\)

<=> \(x=\dfrac{\left(4-x\right)^2}{9}\)

<=> \(\dfrac{9x}{9}=\dfrac{\left(4-x\right)^2}{9}\)

<=> 9x = (4 - x)²

<=> 9x = 16 - 8x + x²

<=> x² - 8x - 9x + 16 = 0

<=> x² - 17x + 16 = 0

<=> x² + x + 16x + 16 = 0

<=> x(x + 1) + 16(x + 1) = 0

<=> (x + 16)(x + 1) = 0

<=> \(\left[{}\begin{matrix}x+16=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-16\\x=-1\end{matrix}\right.\)