Đáp án:a/ $$[_{x = {\text{}} - 1}^{x = 2}$$
b/x=3 hoặc x=1/5
2b/ $$(x + 4)(x - 1)$$
Giải thích các bước giải:a/ $$x\left( {x - 2} \right) + x - 2 = 0 < = > \left( {x - 2} \right)\left( {x + 1} \right) = 0 < = > [_{x + 1 = 0}^{x - 2 = 0} < = > [_{x = {\text{}} - 1}^{x = 2}$$
c/$$\eqalign{
& 5x\left( {x - 3} \right) - x + 3 = 0 \cr
& < = > 5x\left( {x - 3} \right) - \left( {x - 3} \right) = 0 \cr
& < = > (x - 3)(5x - 1) = 0 \cr
& < = > [_{5x - 1 = 0}^{x - 3 = 0} < = > [_{x = \frac{1}{5}}^{x = 3} \cr} $$
2b/Đk: x$\neq$ -1
$$\frac{{{x^3} + 4{x^2} - x - 4}}{{x + 1}} = \frac{{{x^2}(x + 4) - (x + 4)}}{{x + 1}} = \frac{{(x + 4)({x^2} - 1)}}{{x + 1}} = \frac{{(x + 4)(x + 1)(x - 1)}}{{(x + 1)}} = (x + 4)(x - 1)$$
