Câu hỏi của Gin

Question

Tìm x thỏa mãn : ($\sqrt{x}-4$ ).(| x + 2 |-1).(x3-3 ) = 0

Trần Nguyễn Bảo Quyên · Accepted Answer

$\left(\sqrt{x}-4\right)\left(\left|x+2\right|-1\right)\left(x^3-3\right)=0$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}-4=0\\left|x+2\right|-1=0\x^3-3=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}=4\\left|x+2\right|=1\x^3=3\end{matrix}\right.$

$\Rightarrow$$\left[{}\begin{matrix}x=16\\left[{}\begin{matrix}x+2=1\x+2=-1\end{matrix}\right.\x=\sqrt[3]{3}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=16\x=-1\x=-3\x=\sqrt[3]{3}\end{matrix}\right.$

Vậy : ....

#trannguyenbaoquyenCâu trả lời: $\left(\sqrt{x}-4\right)\left(\left|x+2\right|-1\right)\left(x^3-3\right)=0$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}-4=0\\left|x+2\right|-1=0\x^3-3=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}\sqrt{x}=4\\left|x+2\right|=1\x^3=3\end{matrix}\right.$

$\Rightarrow$$\left[{}\begin{matrix}x=16\\left[{}\begin{matrix}x+2=1\x+2=-1\end{matrix}\right.\x=\sqrt[3]{3}\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=16\x=-1\x=-3\x=\sqrt[3]{3}\end{matrix}\right.$

Vậy : ....

#trannguyenbaoquyen

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