g) \(\left\{{}\begin{matrix}x-1\ge0\\2-\sqrt{x-1}\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\sqrt{x-1}\le2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-4\le x-1\le4\end{matrix}\right.\)

\(\Leftrightarrow1\le x\le5\)

h) \(\left\{{}\begin{matrix}\dfrac{2x-4}{5-x}\ge0\\5-x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-4\ge0\\5-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-4\le0\\5-x< 0\end{matrix}\right.\end{matrix}\right.\\x\ne5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}5>x\ge2\left(tm\right)\\5< x\le2\left(vl\right)\end{matrix}\right.\\x\ne5\end{matrix}\right.\)

\(\Leftrightarrow5>x\ge2\)

i) \(x^2-8x-9\ge0\)\(\Leftrightarrow\left(x-4\right)^2-25\ge0\Leftrightarrow\left(x-4\right)^2\ge25\)

\(\Leftrightarrow-5\ge x-4\ge5\)\(\Leftrightarrow-1\ge x\ge9\)

j) \(2x-x^2>0\)

\(\Leftrightarrow\left(x-1\right)^2< 1\)

\(\Leftrightarrow-1< x-1< 1\Leftrightarrow0< x< 2\)

a: ĐKXĐ: \(2\le x\le4\)

b: ĐKXĐ: x>0

c: ĐKXĐ: \(x< \dfrac{1}{3}\)

g: ĐKXĐ: \(1\le x\le4\)

h: ĐKXĐ: \(2\le x< 5\)

i: ĐKXĐ: \(\left[{}\begin{matrix}x\ge9\\x\le-1\end{matrix}\right.\)

\(0< x< 2\)

 chịu

Bài III:

1: Ta có: \(\sqrt{x-3}=5\)

\(\Leftrightarrow x-3=25\)

hay x=28

2: Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}-5}=\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-6=\sqrt{x}-5\)

\(\Leftrightarrow2\sqrt{x}=1\)

hay \(x=\dfrac{1}{4}\)

Câu 4: 

a: Ta có: \(A=\left(4x+1\right)\left(4x-1\right)+2\left(x-1\right)^2-2\left(3x-2\right)^2\)

\(=16x^2-1+2x^2-4x+2-2\left(9x^2-12x+4\right)\)

\(=18x^2-4x+1-18x^2+24x-8\)

\(=20x-7\)

b: Để A=3 thì 20x-7=3

hay \(x=\dfrac{1}{2}\)

Câu 5:

a: Ta có: \(AM=MB=\dfrac{AB}{2}\)

\(AD=BC=\dfrac{AB}{2}\)

\(DN=NC=\dfrac{DC}{2}\)

mà AB=DC

nên AM=MB=CB=NC=ND=AD

Xét ΔAND có AD=DN

nên ΔAND cân tại D

had noticed - hadn't been - had felt - had had - had forgotten - didn't believe - thought - was - had skipped - would - have - wasn't going to tell - could ask - would tell

\(2x^2-16=0\)

\(\Leftrightarrow2x^2=16\)

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow x=\pm\sqrt{8}\)

`2x^2-16=0`

`<=>2x^2=0+16`

`<=>2x^2=16`

`<=>x^2=16:2`

`<=>x^2=8`

`<=>x=+-`\(\sqrt{\text{8}}\)