Câu hỏi của Thuận Minh GilenChi

Question

Tìm x biết :a) $\sqrt{x}$ = $x$

b) $x-2\sqrt{x}=0$

c) $\sqrt{x+1}=1-x$

d) $\sqrt{3x^2+4}+\sqrt{2004x^2+1}=3-4x^2$

e) $\sqrt{3x^2+4}+\sqrt{2007x^2+25}=7-69x^2$

Mashiro Shiina · Accepted Answer

$\sqrt{x}=x$

$\Rightarrow x-\sqrt{x}=0$

$\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\sqrt{x}-1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=0\\left[{}\begin{matrix}x=1\x=-1\end{matrix}\right.\end{matrix}\right.$

$x-2\sqrt{x}=0$

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

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\sqrt{x}-2=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=0\\left[{}\begin{matrix}x=\sqrt{2}\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.$

$\sqrt{x+1}=1-x$

$\Rightarrow\left|x+1\right|=1-2x+x^2$

Với $x\ge-1$ ta có:

$x+1=1-2x+x^2$

$\Rightarrow x+1-1+2x-x^2=0$

$\Rightarrow3x-x^2=0$

$\Rightarrow x\left(3-x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=0\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\x=3\end{matrix}\right.$

Với $x< -1$ ta có:

$-x-1=1-2x+x^2$

$\Rightarrow1-2x+x^2+x-1=0$

$\Rightarrow3x+x^2=0$

$\Rightarrow x\left(3+x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=0\3+x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=0\x=-3\end{matrix}\right.$

Còn pt vô tỉ tui chưa họcCâu trả lời: $\sqrt{x}=x$