Đáp án:
$1) (2x-3)(3x-2)\\
2) (3x+2)(2x+3)\\
3) 3(2x+1)(x+2)\\
4) 3(x-2)(2x-1)\\
5) 2(3x+1)(x+3)\\
6) 2(x-3)(3x-1)\\
7)6(x+1)^2\\
8)(4x-3)(2x+1)\\
9) (2x-1)(4x+3)\\
10) VN\\
11)(2x-3)(4x+1)\\
12) (4x-1)(2x+3)\\
13) (3-x)(8x+1)\\$
Giải thích các bước giải:
$1) 6x^2-13x+6=6x^2-4x-9x+6=2x(3x-2)-3(3x-2)=(2x-3)(3x-2)\\
2) 6x^2+13x+6=6x^2+4x+9x+6=2x(3x+2)+3(3x+2)=(3x+2)(2x+3)\\
3) 6x^2+15x+6=3(2x^2+5x+2)=3(2x^2+4x+x+2)=3\left [ 2x(x+2)+(x+2) \right ]=3(2x+1)(x+2)\\
4) 6x^2-15x+6=3(2x^2-5x+3)=2(2x^2-x-4x+2)=3\left [ x(2x-1)-2(2x-1) \right ]=3(x-2)(2x-1)\\
5) 6x^2 +20x +6=2(3x^2+10x+3)=2(3x^2+9x+x+3)=2\left [ 3x(x+3)+(x+3) \right ]=2(3x+1)(x+3)\\
6) 6x^2 −20x +6=2(3x^2-10x+3)=2(3x^2-x-9x+3)=2\left [ x(3x-1)-3(3x-1) \right ]=2(x-3)(3x-1)\\
7)6x^2 +12x +6=6(x^2+2x+1)=6(x+1)^2\\
8)8x^2 −2x −3=8x^2+4x-6x-3=4x(2x+1)-3(2x+1)=(4x-3)(2x+1)\\
9) 8x^2 +2x −3=8x^2+6x-4x-3=2x(4x+3)-(4x+3)=(2x-1)(4x+3)\\
10) −8x^2+5x −3=-(8x^2-5x+3)=-\left [(2\sqrt{2}x)^2 -2.2\sqrt{2}x.\dfrac{5\sqrt{2}}{8} +\left ( \dfrac{5\sqrt{2}}{8} \right )^2+\dfrac{71}{32}\right ] \\
=-\left [\left ( 2\sqrt{2}x-\dfrac{5\sqrt{2}}{8} \right )^2+\dfrac{71}{32} \right ](VN)$
Vì $\left ( 2\sqrt{2}x-\dfrac{5\sqrt{2}}{8} \right )^2>0\Rightarrow \left [\left ( 2\sqrt{2}x-\dfrac{5\sqrt{2}}{8} \right )^2+\dfrac{71}{32} \right ]>0$
$11)8x^2 −10x −3=8x^2+2x-12x-3=2x(4x+1)-3(4x+1)=(2x-3)(4x+1)\\
12) 8x^2+10x −3=8x^2+12x-2x-3=4x(2x+3)-(2x+3)=(4x-1)(2x+3)\\
13) −8x^2+23x +3=-8x^2-x+24x+3=-x(8x+1)+3(8x+1)=(3-x)(8x+1)\\$
