Question

tìm x,biết

a, \(x-\sqrt{x}-6=0\)

b, \(x+\sqrt{x-12}=0\)

c, \(\sqrt{x+2\sqrt{x-1}=2}\)

d, \(\sqrt{x}-4\sqrt{x}+4=0\)

Answer 1

a) \(x-\sqrt{x}-6=0\) (1)

\(\Leftrightarrow x+2\sqrt{x}-3\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow x=9\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{9\right\}\)

Answer 2

a) \(x-\sqrt{x}-6=0\Leftrightarrow x+2\sqrt{x}-3\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)=0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)

\(\left\{{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\) vậy \(x=9\)

mấy câu sau hình như đề sai