Odpowiedź:
[tex]a)\ \ (3x^2-7x+5)-(9x+6-x^2)=3x^2-7x+5-9x-6+x^2=4x^2-16x-1\\\\b)\ \ 4x(y-3)-3y(2x+y)=4xy-12x-6xy-3y^2=-2xy-12x-3y^2\\\\c)\ \ (6a-2b)(b+7a)=6ab+42a^2-2b^2-14ab=-8ab+42a^2-2b^2\\\\\\d)\ \ \dfrac{4x-6}{2}-\dfrac{9-12x}{3}=\dfrac{\not2(2x-3)}{\not2}-\dfrac{\not3(3-4x)}{\not3}=2x-3-(3-4x)=\\\\\\=2x-3-3+4x=6x-6[/tex]
