Question

Giải phương trình

a) \(\sqrt{9x}=15\) b) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)

Answer 1

Câu a : ĐK : \(x\ge0\)

\(\sqrt{9x}=15\Leftrightarrow9x=225\Leftrightarrow x=\dfrac{225}{9}=25\) ( Thỏa mãn )

Vậy \(S=\left\{25\right\}\)

Câu b : \(x\ge2\)

\(\sqrt{x^2-4}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

Answer 2

a) \(\sqrt{9x}=15\)

<=> \(\sqrt{3^2x}=15\)

<=> \(3\sqrt{x}=15\)

<=> \(\sqrt{x}=5\)

<=> x=5

Answer 3

\(\sqrt{x^2-4}-\sqrt{x-2}\)=0

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

<=> \(\sqrt{x-2}\left(\sqrt{x+2}+1\right)\) = 0

<=> \(\left\{{}\begin{matrix}\sqrt{x+2}=0\\\sqrt{x+2}+1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-2\\\sqrt{x+2}=-1\end{matrix}\right.\)(loại TH2)

Vậy x = -2