a) \(\sqrt[]{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)\(\Leftrightarrow\sqrt[]{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-\left(x^2+2x+1\right)+5\)\(\Leftrightarrow\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(1\right)\)Ta có :\(\left\{{}\begin{matrix}\sqrt[]{3\left(x+1\right)^2+4}\ge2,\forall x\in R\\\sqrt[]{5\left(x+1\right)^2+9}\ge3,\forall x\in R\end{matrix}\right.\)\(\Rightarrow VT=\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge5,\forall x\in R\)\(VP=-\left(x+1\right)^2+5\le5,\forall x\in R\)Dấu "=" xảy ra thì \(VT=VP=5\)\(\left(1\right)\Leftrightarrow\left(x+1\right)^2=0\)\(\Leftrightarrow x+1=0\)\(\Leftrightarrow x=-1\)Vậy nghiệm của phương trình đã cho là \(x=-1\)Câu trả lời: a) \(\sqrt[]{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)\(\Leftrightarrow\sqrt[]{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-\left(x^2+2x+1\right)+5\)\(\Leftrightarrow\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(1\right)\)Ta có :\(\left\{{}\begin{matrix}\sqrt[]{3\left(x+1\right)^2+4}\ge2,\forall x\in R\\\sqrt[]{5\left(x+1\right)^2+9}\ge3,\forall x\in R\end{matrix}\right.\)\(\Rightarrow VT=\sqrt[]{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge5,\forall x\in R\)\(VP=-\left(x+1\right)^2+5\le5,\forall x\in R\)Dấu "=" xảy ra thì \(VT=VP=5\)\(\left(1\right)\Leftrightarrow\left(x+1\right)^2=0\)\(\Leftrightarrow x+1=0\)\(\Leftrightarrow x=-1\)Vậy nghiệm của phương trình đã cho là \(x=-1\)