a,\(\left(a\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x>-2\\x+1< x^2+4x+4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+3x+3>0\end{matrix}\right.\)
Vì \(x^2+3x+3>0\forall x\in R\) (Kết hợp ĐK)
Vậy \(x\ge-1\)
b,\(\left(b\right)\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 6\\x^2-6x+9\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge6\\x^2-6x+9>x^2-12x+36\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x< 6\\\left\{{}\begin{matrix}x\ge6\\6x>27\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 6\\x\ge6\end{matrix}\right.\) \(\Rightarrow T=R\)
c,\(\left(c\right)\Leftrightarrow\sqrt{\left(x-1\right)^2}+3\sqrt{\left(x-5\right)^2}=1\)
\(\Leftrightarrow\left|x-1\right|+3\left|x-5\right|=1\)
Đến đây bạn xét khoảng nhé.
d, \(\left\{{}\begin{matrix}4x^2-8x+3\ge0\\9x^2-6x+1\ge0\end{matrix}\right.\) (*)
\(\left(d\right)\Leftrightarrow\sqrt{4x^2-8x+3}=\sqrt{9x^2-6x+1}\)
\(\Leftrightarrow4x^2-8x+3=9x^2-6x+1\)
\(\Leftrightarrow5x^2+2x-2=0\)
\(\Leftrightarrow x=\frac{-1\pm\sqrt{11}}{5}\) (tm)
Vậy...
