Question

решите уравнение
1) x³-6x²=6-x
2) y³+3y²-4y-12=0
3) 2x³-x²-18x+9=0
4) 4y³-3y²-4y+3=0
5) 2x³-x²-32x+16=0
6) (y+6)²-(y+5)(y-5)=79
пожалуйста, помогите, добрые люди
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Answer 1

Ответ:

1)$x^{3} -6x^{2}=6-x\\x^{3}+x=6x^{2}+6\\x(x^{2}+1)=6(x^{2}+1)\\x=6$

2)$y^{3}+3y^{2}-4y-12=0\\y^{2}(y+3)-4(y+3)=0\\y^{2}(y+3)=4(y+3)\\y^{2}=4\\\left \{ {{y=2} \atop {y=-2}} \right.$

3)$2x^{3}-x^{2}-18x+9=0\\x^{2}(2x-1)-9(2x-1)=0\\x^{2}(2x-1)=9(2x-1)\\x^2=9\\\left \{ {{x=3} \atop {x-3}} \right.$

4)$4y^{3}-3y^{2}-4y+3=0\\4y^3-4y-3y^2+3=0\\4y(y^{2}-1)-3(y^{2}-1)=0\\4y(y^{2}-1)=3(y^{2}-1)\\4y=3\\y=\frac{3}{4}=0.75$

5)$2x^{3}-x^{2}-32x+16=0\\x^{2}(2x-1)-16(2x-1)=0\\x^{2}(2x-1)=16(2x-1)\\x^{2}=16\\\left \{ {{x=4} \atop {x=-4}} \right.$

6)$(y+6)^{2}-(y+5)(y-5)=79\\y^{2}+12y+36-(y^{2}-25)=79\\y^{2}+12y+36-y^{2}+25=79\\y^{2}-y^{2}+12y+36+25=79\\12y=79-61\\12y=18\\y=\frac{18}{12}=\frac{3}{2}=1.5$