1.
[tex]3x(3x - 4) - (3x - 2)(x+5) +4 \\ =9x^2-12x-(3x^2+15x-2x-10)+4 \\ =9x^2-12x-3x^2-15x+2x+10+4 \\ =6x^2-25x+14 \\ \\ (3x-2)(2x-7) \\ =6x^2-21x-4x+14 \\ =6x^2-25x+14 \\ \\ Donc \
3x(3x - 4) - (3x - 2)(x+5) +4 =(3x - 2)(2x - 7)
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2.
[tex]3x(3x- 4) - (3x- 2)(x+5) +4 = 0 \\ 6x^2-25x+14=0 \\ \\ D=b^2-4ac \\ D=(-25)^2-4\times6\times14 \\ D=289 \\ \\ D>0 \\ \\ x_1_,_2= \frac{-b \pm \sqrt{D} }{2a} \\ \\ x_1= \frac{25-17}{2\times12} \boxed{= \frac{2}{3}} \\ \\ \\ x_2= \frac{25+17}{2\times12} \boxed{= \frac{7}{2} }[/tex]
3.
[tex] 3x(3x- 4) - (3x- 2)(x+5) +4 = 14 \\ 6x^2-25x+14=14 \\ 6x^2-25x=0 \\ x(6x-25)=0 \\ \boxed{x=0} \\ \\ 6x-25=0 \\ 6x=25 \\\boxed{ x=25/6}[/tex]
Si tu as des questions, n'hésite pas! =)
