Cho : $\sqrt[]{x}$ $\geq$ 0 , x $\neq$ 9 giải bất pt sau : ($\sqrt[]{x}$ – 2$)^{2}$ $\leq$ 2 29/09/2021 Bởi Harper Cho : $\sqrt[]{x}$ $\geq$ 0 , x $\neq$ 9 giải bất pt sau : ($\sqrt[]{x}$ – 2$)^{2}$ $\leq$ 2
($\sqrt[]{x}$-2)$^{2}$ $\leq$ 2 ⇔\(\left[ \begin{array}{l}\sqrt[]{x}-2\leq\sqrt[]{2}\\-\sqrt[]{x}+2\leq\sqrt[]{2}\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x\leq\sqrt[]{2}+2\\x\leq\sqrt[]{2}-2\end{array} \right.\) Vậy 0$\leq$x$\leq$ $\sqrt[]{2}+2$ hoặc $\leq$ $\sqrt[]{2}-2$ Bình luận
$⇔ |\sqrt[]{x}-2|≤\sqrt[]{2}$ $⇔\sqrt[]{x}-2≤\sqrt[]{2}(1)$ hoặc $2-\sqrt[]{x}≤\sqrt[]{2}(2)$ $(1)⇔\sqrt[]{x}≤\sqrt[]{2}+2⇔x≤(\sqrt[]{2}+2)^2$ $(2)⇔-\sqrt[]{x}≤\sqrt[]{2}-2⇔\sqrt[]{x}≥2-\sqrt[]{2}⇔x≥(2-\sqrt[]{2})^2$ Vậy $0≤x≤(\sqrt[]{2}+2)^2$ hoặc $x≥(2-\sqrt[]{2})^2$ Bình luận
($\sqrt[]{x}$-2)$^{2}$ $\leq$ 2
⇔\(\left[ \begin{array}{l}\sqrt[]{x}-2\leq\sqrt[]{2}\\-\sqrt[]{x}+2\leq\sqrt[]{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x\leq\sqrt[]{2}+2\\x\leq\sqrt[]{2}-2\end{array} \right.\)
Vậy 0$\leq$x$\leq$ $\sqrt[]{2}+2$ hoặc $\leq$ $\sqrt[]{2}-2$
$⇔ |\sqrt[]{x}-2|≤\sqrt[]{2}$
$⇔\sqrt[]{x}-2≤\sqrt[]{2}(1)$ hoặc $2-\sqrt[]{x}≤\sqrt[]{2}(2)$
$(1)⇔\sqrt[]{x}≤\sqrt[]{2}+2⇔x≤(\sqrt[]{2}+2)^2$
$(2)⇔-\sqrt[]{x}≤\sqrt[]{2}-2⇔\sqrt[]{x}≥2-\sqrt[]{2}⇔x≥(2-\sqrt[]{2})^2$
Vậy $0≤x≤(\sqrt[]{2}+2)^2$ hoặc $x≥(2-\sqrt[]{2})^2$