Đáp án:
2
Giải thích các bước giải:
$$\eqalign{
& y = {{{x^2} - x + 1} \over x} = x - 1 + {1 \over x}\,\,\left( {x \ne 0} \right) \cr
& \Rightarrow y' = 1 - {1 \over {{x^2}}} \cr
& Goi\,\,dt\,\,di\,\,qua\,\,M\left( {2;2} \right)\,\,la\,\,y = k\left( {x - 2} \right) + 2\,\,\left( d \right) \cr
& \left( d \right)\,\,tiep\,\,xuc\,\,voi\,\,y = {{{x^2} - x + 1} \over x} \cr
& \Leftrightarrow \left\{ \matrix{
{{{x^2} - x + 1} \over x} = k\left( {x - 2} \right) + 2\,\,\left( 1 \right) \hfill \cr
k = 1 - {1 \over {{x^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right) \hfill \cr} \right. \cr
& Thay\,\,\left( 2 \right)\,\,vao\,\,\left( 1 \right) \cr
& \Rightarrow {{{x^2} - x + 1} \over x} = \left( {1 - {1 \over {{x^2}}}} \right)\left( {x - 2} \right) + 2 \cr
& \Leftrightarrow {{{x^2} - x + 1} \over x} = {{\left( {{x^2} - 1} \right)\left( {x - 2} \right) + 2{x^2}} \over {{x^2}}} \cr
& \Leftrightarrow {x^3} - {x^2} + x = {x^3} - 2{x^2} - x + 2 + 2{x^2} \cr
& \Leftrightarrow - {x^2} + 2x - 2 = 0\,\,\left( * \right) \cr
& PT\,\,\left( * \right)\,\,co\,\,2\,\,nghiem\,\,pb\, \ne 0 \cr
& \Rightarrow Tu\,\,M\,\,ke\,\,duoc\,\,2\,\,tiep\,\,\,tuyen. \cr} $$
