Đáp án + giải thích các bước giải:
`lim_{x->1}(\sqrt{4x}+\sqrt{3x+1}-4)/(x^2-1)=lim_{x->1}(\sqrt{4x}-2+\sqrt{3x+1}-2)/(x^2-1)=lim_{x->1}[(4x-4)/((x-1)(x+1)(\sqrt{4x}+2))+(3x+1-4)/((x-1)(x+1)(\sqrt{3x+1}+2))]=lim_{x->1}[4/((x+1)(\sqrt{4x}+2))+3/((x+1)(\sqrt{3x+1}+2))]=4/((1+1)(\sqrt{4.1}+2))+3/((1+1)(\sqrt{3.1+1}+2))=7/8`
