Đề đúng là +2 trên tử phải nằm trong căn đầu tiên, nếu ko giới hạn sẽ là dương vô cùng
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}+\frac{8-\left(7x+1\right)}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{x-1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}}\)
\(=\frac{\frac{3}{4}-\frac{1}{4+4+4}}{\sqrt{2}}=\frac{2}{3\sqrt{2}}=\frac{\sqrt{2}}{3}+0\)
\(\Rightarrow a+b+c=1+3+0=4\)
