giải bất pt 1)16-x^2>hoặc =0 2)x^2-x+1>0 02/09/2021 Bởi Savannah giải bất pt 1)16-x^2>hoặc =0 2)x^2-x+1>0
$\begin{array}{l}1)\,\, 16-x^2≥0\\-x^2≥-16\\x^2≤16\\\left[ \begin{array}{l}x\ge-4\\x\le4\end{array} \right.\\-4≤x≤4\\2)\,\, x^2-x+1>0\\x^2-x+\dfrac{1}{4}+\dfrac{3}{4}>0\\(x-\dfrac{1}{2})^2+\dfrac{3}{4}>0∀x\end{array}$ Bình luận
1) `16-x^2≥0` `⇔x^2≤16` `⇔`\(\left[ \begin{array}{l}x\ge-4\\x\le4\end{array} \right.\) `⇔-4≤x≤4` 2) `x^2-x+1>0` `⇔x^2-x+1/4+3/4>0` `⇔(x-1/2)^2+3/4>0` (luôn đúng) `⇔x∈R` Bình luận
$\begin{array}{l}1)\,\, 16-x^2≥0\\-x^2≥-16\\x^2≤16\\\left[ \begin{array}{l}x\ge-4\\x\le4\end{array} \right.\\-4≤x≤4\\2)\,\, x^2-x+1>0\\x^2-x+\dfrac{1}{4}+\dfrac{3}{4}>0\\(x-\dfrac{1}{2})^2+\dfrac{3}{4}>0∀x\end{array}$
1)
`16-x^2≥0`
`⇔x^2≤16`
`⇔`\(\left[ \begin{array}{l}x\ge-4\\x\le4\end{array} \right.\)
`⇔-4≤x≤4`
2)
`x^2-x+1>0`
`⇔x^2-x+1/4+3/4>0`
`⇔(x-1/2)^2+3/4>0` (luôn đúng)
`⇔x∈R`