`[12x+1]/12 >= [9x+3]/3-[8x+1]/4`

`<=>12x+1 >= 4(9x+3)-3(8x+1)`

`<=>12x+1 >= 36x+12-24x-3`

`<=>12x-36x+24x >= 12-3-1`

`<=>0 >= 8` (Vô lí)

Vậy bất ptr vô nghiệm.

Ta có: \(6-\left(x-\dfrac{1}{3}\right)^2=2^{2021}:\left(-2\right)^{2020}\)

\(\Leftrightarrow6-\left(x-\dfrac{1}{3}\right)^2=2^{2021}:2^{2020}\\ \Leftrightarrow6-\left(x-\dfrac{1}{3}\right)^2=2\\ \Leftrightarrow\left(x-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=2\\x-\dfrac{1}{3}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

    Vậy \(\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

a oval face => an oval face

likes dance to music => likes dancing 

we are => we will be

She has a oval face ->. She has an oval face

mik thấy ở dòng cuối sai từ bkie 

bn đăng lại đi chúng mk giải cho

tui chịu luôn đó

Câu 1 này bình phương hơi dài dòng nên khử trị tuyệt đối sẽ tốt hơn:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-2x-3\ge0\\x^2-2x-3\ge3x-3\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-2x-3< 0\\-\left(x^2-2x-3\right)\ge3x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge3\\x\le-1\end{matrix}\right.\\x^2-5x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}-1< x< 3\\x^2+x-6\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge3\\x\le-1\end{matrix}\right.\\\left[{}\begin{matrix}x\ge5\\x\le0\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}-1< x< 3\\-3\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\\-1< x\le2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)

Vậy C đúng

12.D thành A
13.D
14.B
Còn lại đúng rồi ạ