a) ĐK x>2

a) Để giá trị của biểu thức \(\frac{x}{x^2-4}+\sqrt{x-2}\)xác định được thì

\(\left\{{}\begin{matrix}x^2-4\ne0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\notin\left\{2;-2\right\}\\x\ge2\end{matrix}\right.\Leftrightarrow x>2\)

b) Để giá trị của biểu thức \(\frac{\sqrt{x}}{\left|x\right|-1}\) xác định được thì

\(\left\{{}\begin{matrix}x\ge0\\\left|x\right|-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left|x\right|\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\notin\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow0\le x\ne1\)

Lời giải:

a) Ta thấy $2x^2+3>0$ với mọi $x\in\mathbb{R}$ nên $\frac{-4}{2x^2+3}< 0$ với mọi $x\in\mathbb{R}$

Do đó biểu thức không có nghĩa với mọi $x\in\mathbb{R}$, hay không tồn tại giá trị $x$ để bt có nghĩa

b)

ĐK: \(\left\{\begin{matrix} x^2+2\neq 0\\ -\frac{1-x}{x^2+2}\geq 0\end{matrix}\right.\Leftrightarrow -(1-x)\geq 0\Leftrightarrow 1-x\leq 0\Leftrightarrow x\geq 1\)

c) ĐK: $-x^2+2x+1\geq 0$

$\Leftrightarrow x^2-2x-1\leq 0$

$\Leftrightarrow (x-1)^2\leq 2\Leftrightarrow -\sqrt{2}\leq x-1\leq \sqrt{2}$

$\Leftrightarrow 1-\sqrt{2}\leq x\leq 1+\sqrt{2}$

d)

ĐK: $4-|x|\geq 0\Leftrightarrow |x|\leq 4\Leftrightarrow -4\leq x\leq 4$

e)

ĐK: $x^2-16>0\Leftrightarrow (x-4)(x+4)>0\Leftrightarrow x>4$ hoặc $x< -4$

a, \(x+1\ge0\Leftrightarrow x\ge-1\)

b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)

d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)

e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)

f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)

g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:

\(\left(x-3\right)\left(7-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow3\le x\le7\)

c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:

\(\left(2x-5\right)\left(2x+5\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow x>2\)

1: ĐKXĐ: 2-3x>=0

=>x<=2/3

2: ĐKXĐ: -3x^2>=0

=>x^2<=0

=>x=0

3: ĐKXĐ: -2023x^3>=0

=>x^3<=0

=>x<=0

4: ĐKXĐ: -2(x-5)>=0

=>x-5<=0

=>x<=5

5: ĐKXĐ: -5/2-2x>=0

=>2-2x<0

=>2x>2

=>x>1

6: ĐKXĐ: (x^2+1)(3-2x)>=0

=>3-2x>=0

=>-2x>=-3

=>x<=3/2

7: ĐKXĐ: (-x^2-1)(3-x)>=0

=>(x^2+1)(x-3)>=0

=>x-3>=0

=>x>=3

1) a) x<=11/2

b) x>=2

c) x#0

d) x>7

\(1,\\ a,ĐK:11-2x\ge0\Leftrightarrow x\le\dfrac{11}{2}\\ b,ĐK:9x-18\ge0\Leftrightarrow x\ge2\\ c,ĐK:x\ne0;\dfrac{3}{x^2}\ge0\left(luôn.đúng.do.3>0;x^2>0\right)\Leftrightarrow x\in R\backslash\left\{0\right\}\\ d,ĐK:\dfrac{5}{x-7}\ge0\Leftrightarrow x-7>0\left(5>0;x-7\ne0\right)\Leftrightarrow x>7\\ 2,\\ a,=\left|4x\right|-2x^2=4x-2x^2\\ b,bạn.thiếu.điều.kiện.nhé\\ c,=\left|x-5\right|-4x=5-x-4x=5-5x\)

Bài 2:

a: \(\sqrt{16x^2}-2x^2=4x-2x^2\)

c: \(\sqrt{\left(x-5\right)^2}-4x=5-4x-x=5-5x\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.