Câu hỏi của ??????

a.\(M=16x^2+24x+9-2x^2-12x-5\left(x^2-4\right)\)\(=9x^2+12x+29\)b.\(x=-2\Rightarrow=9.\left(-2\right)^2+12.\left(-2\right)+29=41\)c.\(M=29\Rightarrow9x^2+12x+29=29\)\(\Leftrightarrow9x^2+12x=0\)\(\Leftrightarrow3x\left(3x+4\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{4}{3}\end{matrix}\right.\)d.\(M=9x^2+12x+29=\left(3x+2\right)^2+25\)Do \(\left\{{}\begin{matrix}\left(3x+2\right)^2\ge0\\25>0\end{matrix}\right.\) với mọi x \(\Rightarrow m=\left(3x+2\right)^2+25>0;\forall x\)Câu trả lời: a.\(M=16x^2+24x+9-2x^2-12x-5\left(x^2-4\right)\)\(=9x^2+12x+29\)b.\(x=-2\Rightarrow=9.\left(-2\right)^2+12.\left(-2\right)+29=41\)c.\(M=29\Rightarrow9x^2+12x+29=29\)\(\Leftrightarrow9x^2+12x=0\)\(\Leftrightarrow3x\left(3x+4\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{4}{3}\end{matrix}\right.\)d.\(M=9x^2+12x+29=\left(3x+2\right)^2+25\)Do \(\left\{{}\begin{matrix}\left(3x+2\right)^2\ge0\\25>0\end{matrix}\right.\) với mọi x \(\Rightarrow m=\left(3x+2\right)^2+25>0;\forall x\)