Question

có ai bt giải bài nàykhông giúp mk vs mk đg cần rất rất gấp mong các bn giúp cho ( xin cảm ơn) 

Answer 1

Bài nào?

Answer 2

5.

\(A=B\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x>-2\end{matrix}\right.\Leftrightarrow x\ge5\)

Answer 3

6.

a, ĐK: \(\left\{{}\begin{matrix}2-\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\le2\\x\ge1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ge1\end{matrix}\right.\Leftrightarrow1\le x\le5\)

b, ĐK: \(\left\{{}\begin{matrix}x-2\ge0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne2\end{matrix}\right.\Leftrightarrow x>2\)

c, ĐK: \(\left\{{}\begin{matrix}x^2-4\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x>2\end{matrix}\right.\Leftrightarrow x>2\)

d, ĐK: \(\left\{{}\begin{matrix}6x-1\ge0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{6}\\x\ge-3\end{matrix}\right.\Leftrightarrow x\ge\dfrac{1}{6}\)

Answer 4

7.

a, ĐK: \(\left\{{}\begin{matrix}\dfrac{3}{3x+2}\ge0\\3x+2\ne0\end{matrix}\right.\Leftrightarrow3x+2>0\Leftrightarrow x>-\dfrac{2}{3}\)

b, ĐK: \(\left\{{}\begin{matrix}-\dfrac{3}{2x+6}\ge0\\2x+6\ne0\end{matrix}\right.\Leftrightarrow2x+6< 0\Leftrightarrow x< -3\)

c, ĐK: \(\left\{{}\begin{matrix}3-x\ge0\\7x+2>0\end{matrix}\right.\Leftrightarrow-\dfrac{2}{7}< x\le3\)

d, ĐK: \(x\ne0\)