a) ĐKXĐ: \(-x-8\ge0\Leftrightarrow x\le-8\)

b) ĐKXĐ: \(x^2-2x+1>0\Leftrightarrow\left(x-1\right)^2>0\Leftrightarrow x\ne1\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\5-x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)

d) ĐKXĐ: \(x^2+3\ge0\left(đúng.do.x^2+3\ge3>0\right)\)

1) 

a) \(\sqrt{2x-4}\) có nghĩa khi:

\(2x-4\ge0\)

\(\Leftrightarrow2x\ge4\)

\(\Leftrightarrow x\ge\dfrac{4}{2}\)

\(\Leftrightarrow x\ge2\)

b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi 

\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)

\(\Rightarrow4-x\le0\)

\(\Leftrightarrow x\ge4\)

2) 

a) \(A=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\cdot\sqrt{5}+2^2}\)

\(A=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(A=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)

\(A=\sqrt{5}+2-\sqrt{5}+2\)

\(A=4\)

\(B=\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-5}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{5}-\sqrt{7}}\)

\(B=\left(-\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(B=\left[-\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=\left(-\sqrt{7}-\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(B=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(B=-\left(7-5\right)\)

\(B=-2\)

\(a,f\left(x\right)=\sqrt{2x-7}\)

\(f\left(x\right)\) có nghĩa \(\Leftrightarrow2x-7\ge0\Leftrightarrow x\ge\dfrac{7}{2}\)

\(b,f\left(x\right)=\sqrt{-3x+4}\)

\(f\left(x\right)\) có nghĩa \(\Leftrightarrow-3x+4\ge0\Leftrightarrow x\le\dfrac{4}{3}\)

\(c,f\left(x\right)=\sqrt{1+x^2}\)

\(f\left(x\right)\) có nghĩa \(\Leftrightarrow1+x^2\ge0\)

Mà \(1+x^2\ge0\) với mọi x \( \left(x^2\ge0\Rightarrow x^2+1\ge0\right)\)

\(\sqrt{1+x^2}\) có nghĩa với mọi x

\(a,ĐK:x\in R\)

\(b,ĐK:\dfrac{-7}{8-10x}\ge0\Leftrightarrow8-10x< 0\left(-7< 0\right)\Leftrightarrow x>\dfrac{4}{5}\)

\(c,ĐK:\dfrac{24-6x}{-7}\ge0\Leftrightarrow24-6x\le0\left(-7< 0\right)\Leftrightarrow x\ge4\)

a:ĐKXĐ: \(x\in R\)

b: ĐKXĐ: \(x>\dfrac{4}{5}\)

c: ĐKXĐ: x>4

a) Để căn thức bậc 2 có nghĩa \(\Rightarrow3-5x\ge0\Rightarrow x\le\dfrac{3}{5}\)

b) Để căn thức bậc 2 có nghĩa \(\Rightarrow\dfrac{5}{2x+1}\ge0\Rightarrow2x+1>0\Rightarrow x>-\dfrac{1}{2}\)

\(a,x\le\dfrac{3}{5}\)

b,\(x>-\dfrac{1}{2}\)

a, để căn thức có nghĩa thì 3-5x≥0⇔x≤\(\dfrac{3}{5}\)

b, để căn thức có nghĩa thì 2x+1>0⇔x>\(\dfrac{-1}{2}\)

Muốn \(\sqrt{\dfrac{1}{x-3}}\) có nghĩa thì cần:

\(x-3\ge0\)

<=> \(x\ge3\)

Vậy ĐKXĐ là: \(x\ge3\)

Thay dấu ''\(\ge\)'' thành dấu ''>'' giùm mik.

a) để biểu thức có nghĩa thì \(\dfrac{2x-8}{x^2+1}\ge0\) mà \(x^2+1>0\)

\(\Rightarrow2x-8\ge0\Rightarrow x\ge4\)

b)  để biểu thức có nghĩa thì \(\dfrac{-x^2-3}{8x+10}\ge0\) mà \(-x^2-3=-\left(x^2+3\right)< 0\)

\(\Rightarrow8x+10< 0\Rightarrow x< -\dfrac{5}{4}\)

c)  để biểu thức có nghĩa thì \(x^2-2x+1>0\Rightarrow\left(x-1\right)^2>0\Rightarrow x\ne1\)

a) ĐKXĐ: \(x\ge4\)

b) ĐKXĐ: \(x< -\dfrac{5}{4}\)

c) ĐKXĐ: \(x\ne1\)

Bài 1 :

a, ĐKXĐ : \(3-2x\ge0\)

\(\Rightarrow x\le\dfrac{3}{2}\)

Vậy ...

b, ĐKXĐ : \(\left\{{}\begin{matrix}-\dfrac{5}{2x+1}\ge0\\2x+1\ne0\end{matrix}\right.\)

\(\Rightarrow2x+1< 0\)

\(\Rightarrow x< -\dfrac{1}{2}\)

Vậy ...

a,ĐKXĐ \(3-2\text{x}>0\Leftrightarrow-2x>-3\Leftrightarrow\text{x}< \dfrac{3}{2}\)

b,\(\dfrac{-5}{2x+1}>0\Leftrightarrow2x+1< 0\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)

( bây giờ mình bận nên làm trước 2 bài =))

a, \(x\le\dfrac{3}{2}\)

b, \(x< -\dfrac{1}{2}\)

*a, \(\sqrt{\left(2x-3\right)^2}=5=>|2x-3|=5=>\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

b, \(\sqrt{9x+9}+\sqrt{4x+4}-\sqrt{16x+16}=3\)

\(< =>3\sqrt{x+1}+2\sqrt{x+1}-4\sqrt{x+1}=3\)\(\left(x\ge-1\right)\)

\(< =>\sqrt{x+1}=3=>x+1=9=>x=8\left(tm\right)\)

a) \(\sqrt{4x^2-16}\)

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

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

để phương trình trên có nghĩa

⇒2x-4≥0

⇒x≥2

a) \(ĐK:4x^2-16\ge0\)

\(\Leftrightarrow4x^2\ge16\Leftrightarrow x^2\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

b) \(ĐK:9x^2-25\ge0\)

\(\Leftrightarrow9x^2\ge25\)\(\Leftrightarrow x^2\ge\dfrac{25}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x\le-\dfrac{5}{3}\end{matrix}\right.\)

\(a,ĐK:4x^2-16\ge0\Leftrightarrow4\left(x-2\right)\left(x+2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

\(b,ĐK:9x^2-25\ge0\Leftrightarrow\left(3x-5\right)\left(3x+5\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\3x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\3x+5\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x\le-\dfrac{5}{3}\end{matrix}\right.\)