Question

x-6 can x+8=0

Answer 1

\(x-6\sqrt{x}+8=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\)

Answer 2

\(x-6\sqrt{x}+8=0\)                  ĐK: \(x\ge0\)

<=> \(x-2\sqrt{x}-4\sqrt{x}+8=0\)

<=> \(\left(\sqrt{x}\right)^2-2\sqrt{x}-4\sqrt{x}+8=0\)

<=> \(\sqrt{x}\left(\sqrt{x}-2\right)-4\left(\sqrt{x}-2\right)=0\)

<=> \(\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x}-4=0\\\sqrt{x}-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=16\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

Answer 3

\(2x-5\sqrt{x}+3\)

= \(2x-6\sqrt{x}+\sqrt{x}+3\)

= \(2.\left(\sqrt{x}\right)^2-6\sqrt{x}+\sqrt{x}+3\)

= \(2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)\)

= \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)