Đáp án:
Giải thích các bước giải:
$Bài4_{}$
⇔ $a)(x-3)(x-1)-3(x-3)_{}$
⇔ $(x-3)(x-1-3)_{}$
⇔ $(x-3)(x-4)_{}$
$b)(x-1)(2x+1)+3(x-1)(x+2)_{}$
⇔ $(x-1)[ 2x+1+3(x+2)]_{}$
⇔ $(x-1)(2x+1+3x+6)_{}$
⇔ $(x-1)(5x+7)_{}$
$c)(6x+3)-(2x-5)(2x+1)_{}$
⇔ $[ 3(2x+1)]-(2x-5)(2x+1)_{}$
⇔ $3(2x+1)-(2x-5)(2x+1)_{}$
⇔ $(2x+1)[ 3-(2x+5)]_{}$
⇔ $(2x+1)(8-2x)_{}$
⇔ $2(2x+1)(4-x)_{}$
$d)(x-5)^2+(x+5)(x-5)-(5-x)_{}$
⇔ $(x-5)^2+(x+5)(x-5)-[ -(x-5)]_{}$
⇔ $(x-5)^2+(x+5)(x-5)+(x-5)_{}$
⇔ $(x-5)(x-5+x+5+1)_{}$
⇔ $(x-5)(2x+1)_{}$
$e)(3x+2)(4x-3)-(2-3x)(x-1)-2(3x-2)(x+1)_{}$
⇔ $12x^2-9x+8x-6-(2x-2-3x^2+3x)+(-6x+4)(x+1)_{}$
⇔ $12x^2-9x+8x-6-(5-2-3x^2)-6x^2-6x+4x+4_{}$
⇔ $12x^2-9x+8x-6-5x+2+3x^2-6x^2-6x+4x+4_{}$
⇔ $9x^2-8x_{}$
⇔ $x(9x-8)_{}$
