Tìm \(m\) để: \[\mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} - 2x + 1} \right)\left( {x - m} \right) - m(x - 2) - 1}}{{2m{x^3} - (4m + 1){x^2} + 2(m + 1) - 1}} = \frac{{20}}{{23}}\]

\[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} - 2x + 1} \right)\left( {x - m} \right) - m(x - 2) - 1}}{{2m{x^3} - (4m + 1){x^2} + 2(m + 1) - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} - x - 1} \right)\left( {x - m} \right) - {x^2} + 2x - 1}}{{(x - 1)(2m{x^2} - (2m + 1)x - 2m - 1)}}\end{array}\]

\[ = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{{{(x - 1)}^2}(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x - m} \right) - {{(x - 1)}^2}}}{{(x - 1)(2m{x^2} - (2m + 1)x - 2m - 1)}} = \mathop {\lim }\limits_{x \to 1} (x - 1)\frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x - m} \right) - 1}}{{(2m{x^2} - (2m + 1)x - 2m - 1)}}\]TH 1: 1 là nghiệm của \[2m{x^2} - (2m + 1)x - 2m - 1 = 0 \Rightarrow m = - 1\]. Khi đó

\[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} (x - 1)\frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x - m} \right) - 1}}{{(2m{x^2} - (2m + 1)x - 2m - 1)}} = \mathop {\lim }\limits_{x \to 1} (x - 1)\frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x + 1} \right) - 1}}{{ - 2{x^2} + x + 1}}\\ = \mathop {\lim }\limits_{x \to 1} (x - 1)\frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x + 1} \right) - 1}}{{(x - 1)( - 2x - 1)}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x + 1} \right) - 1}}{{( - 2x - 1)}} = - \frac{1}{3}\end{array}\] Không thỏa mãn

TH 2: 1 không là nghiệm của \[2m{x^2} - (2m + 1)x - 2m - 1 = 0 \Rightarrow m \ne - 1\]. Khi đó

\[\mathop {\lim }\limits_{x \to 1} (x - 1)\frac{{\frac{{(x + 3)}}{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} + x + 1} \right)}}\left( {x - m} \right) - 1}}{{(2m{x^2} - (2m + 1)x - 2m - 1)}} = 0\](Không thỏa mãn)

Vậy không tồn tại m để \[\mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^3} + 2{x^2} - 3x + 4} - 2x + 1} \right)\left( {x - m} \right) - m(x - 2) - 1}}{{2m{x^3} - (4m + 1){x^2} + 2(m + 1) - 1}} = \frac{{20}}{{23}}\]