Объяснение:

6.

[tex]x^3+4x^2-5x=0\\\\x*(x^2+4x-5)=0\\\\x*(x^2+5x-x-5)=0\\\\x*(x*(x+5)-(x+5))=0\\\\x*(x+5)*(x-1)=0\\\\x_1=0.\\\\x+5=0\\\\x_2=-5.\\\\x-1=0\\\\x_3=1.[/tex]

Ответ: x₁=1, x₂=-5, x₃=1.

7. 1)

[tex]\displaystyle\\\frac{x^2+3x}{x^2+x-6} =\frac{x*(x+3)}{x^2+3x-2x-6} =\\\\=\frac{x*(x+3)}{x*(x+3)-2*(x+3)}=\frac{x*(x+3)}{(x+3)*(x-2)}=\frac{x}{x-2}.[/tex]

7. 2)

[tex]\displaystyle\\\frac{3x^2-7x+2}{x^2-4} =\frac{3x^2-6x-x+2}{x^2-2^2} =\frac{3x*(x-2)-(x-2)}{(x-2)*(x+2)} =\\\\=\frac{(x-2)*(3x-1)}{(x-2)*(x+2)}=\frac{3x-1}{x+2}.[/tex]