ĐKXĐ: x2-8x+14≥0 ⇔ \(x^2-\left(4+\sqrt{2}\right)x-\left(4-\sqrt{2}\right)x+14\)≥0
⇔ \(x\left(x-4-\sqrt{2}\right)-\left(4-\sqrt{2}\right)\left(x-4-\sqrt{2}\right)\)≥0
⇔ \(\left(x-4-\sqrt{2}\right)\left(x-4+\sqrt{2}\right)\)≥0
⇔ {x-4-√2≥0 ⇔ x≥4+√2
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{x-4+√2≥0
⇔ {x-4-√2≤0 ⇔ x≤4-√2
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{x-4+√2≤0
⇔ x≥4+√2, x≤4-√2
Vậy ...
Xét : \(x^2-8x+14\ge0\)
\(\Leftrightarrow x^2-2.x.4+16-2\ge0\)
\(\Leftrightarrow\left(x-4\right)^2-2\ge0\)
\(\Leftrightarrow\left(x-4+\sqrt{2}\right)\left(x-4-\sqrt{2}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4+\sqrt{2}\ge0\\x-4-\sqrt{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4+\sqrt{2}\le0\\x-4-\sqrt{2}\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4-\sqrt{2}\\x\ge4+\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4-\sqrt{2}\\x\le4+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\)
Vậy \(x\ge4+\sqrt{2}\) ; \(x\le4-\sqrt{2}\) thì căn thức đc xác định.
ĐKXĐ: \(\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\)
