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: \(\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\)

Điều kiện xác định của biểu thức là:

\(2x+1>0\) được \(x>-\dfrac{1}{2}\)

\(x^2\le16\) được \(-4\le x\le4\)

\(x^2-8x+14\ge0\)

\(x^2-8x+14\ge0\Leftrightarrow\left(x-4\right)^2\ge2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4\le-\sqrt{2}\\x-4\ge\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le4-\sqrt{2}\\x\ge4+\sqrt{2}\end{matrix}\right.\)

Vậy đkxđ của biểu thức là:

\(-\dfrac{1}{2}< x\le4-\sqrt{2}\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x< -12\end{matrix}\right.\)

a/ \(1-16x^2\ge0\Rightarrow x^2\le16\Rightarrow-\frac{1}{4}\le x\le\frac{1}{4}\)

b/ \(\left\{{}\begin{matrix}x^2-3\ge0\\x^2-3\ne1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\\x\ne\pm2\end{matrix}\right.\)

c/ \(8x-x^2-15\ge0\Rightarrow3\le x\le5\)

d/ Hàm số xác định với mọi x

e/ \(\left\{{}\begin{matrix}x\ge\frac{1}{2}\\x\ne1\end{matrix}\right.\)

f/ \(\left\{{}\begin{matrix}-4\le x\le4\\x>-\frac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow-\frac{1}{2}< x\le4-\sqrt{2}\)

\(a,\)\(\sqrt{\frac{1}{\left(x-3\right)^2}}\)

\(đk:\)\(\frac{1}{\left(x-3\right)^3}\ne0\)\(\Rightarrow\left(x-3\right)^3\ne0\)\(\Leftrightarrow x\ne3\)

Và \(\frac{1}{\left(x-3\right)}>0\Rightarrow x-3>0\)\(\Rightarrow x>3\)

Vậy để căn thức xác định thì x > 3

\(\sqrt{8x-x^2-15}\)

\(=\sqrt{-\left(x^2-8x+15\right)}\)

\(=\sqrt{-\left(x^2-8x+16-1\right)}\)

\(=\sqrt{-\left[\left(x^2-8x+16\right)-1\right]}\)

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

\(đk:\)\(-\left(x-4\right)^2+1\ge0\)

\(\Rightarrow\left(x-4\right)^2\le1\)

\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=1\\\left(x-4\right)^2=0\end{cases}}\)

\(\left(x-4\right)^2=1\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

\(\left(x-4\right)^2=0\Rightarrow x=4\)

Vậy căn thức xác định \(\Leftrightarrow x=\left\{3;4;5\right\}\)

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)