Question

решите пожажда :с...
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Answer 1

$1)(2-3x)(4x+1)=8x+2-12x^2-3x=-12x^2+5x+2\\\\2)(3y-2)(y^2-y+1)=3y^3-3y^2+3y-2y^2+2y-2=3y^3-5y^2+5y-2\\\\3)(5+2y)(y^2+2y-3)=5y^2+10y-15+2y^3+4y^2-6y=2y^3+9y^2+4y-\\-15\\\\4)9y^2-(3y-1)(3y-2)=-16\\9y^2-(9y^2-6y-3y+2)=-16\\9y^2-9y^2+6y+3y-2=-16\\9y=-16+2\\9y=-14\\\\y=-\frac{14}{9}$