![1)x^3+6x^2-x-6=0 \\P(1)=1+6-1-6=0 \Rightarrow x_1=1 \\(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=\\=x^3+x^2(a-1)+x(b-a)-b \\x^3+6x^2-x-6=x^3+x^2(a-1)+x(b-a)-b \\ \left \{ {{a-1=6} \atop {b-a=-1}} \right. \Rightarrow \left \{ {{a=7} \atop {b=6}}\right. \\(x-1)(x^2+7x+6)=0 \\x^2+7x+6=0 \\D=49-24=25=5^2 \\x_2= \frac{-7+5}{2} =-1 \\x_3= \frac{-7-5}{2} =-6](https://tex.z-dn.net/?f=1%29x%5E3%2B6x%5E2-x-6%3D0+%5C%5CP%281%29%3D1%2B6-1-6%3D0+%5CRightarrow+x_1%3D1+%5C%5C%28x-1%29%28x%5E2%2Bax%2Bb%29%3Dx%5E3%2Bax%5E2%2Bbx-x%5E2-ax-b%3D%5C%5C%3Dx%5E3%2Bx%5E2%28a-1%29%2Bx%28b-a%29-b+%5C%5Cx%5E3%2B6x%5E2-x-6%3Dx%5E3%2Bx%5E2%28a-1%29%2Bx%28b-a%29-b+%5C%5C+%5Cleft+%5C%7B+%7B%7Ba-1%3D6%7D+%5Catop+%7Bb-a%3D-1%7D%7D+%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Ba%3D7%7D+%5Catop+%7Bb%3D6%7D%7D%5Cright.+%5C%5C%28x-1%29%28x%5E2%2B7x%2B6%29%3D0+%5C%5Cx%5E2%2B7x%2B6%3D0+%5C%5CD%3D49-24%3D25%3D5%5E2+%5C%5Cx_2%3D+%5Cfrac%7B-7%2B5%7D%7B2%7D+%3D-1+%5C%5Cx_3%3D+%5Cfrac%7B-7-5%7D%7B2%7D+%3D-6)
Ответ: x1=1; x2=-1; x3=-6
![2)x^3-2x^2-9x+18=0 \\P(2)=8-8-18+18=0\Rightarrow x_1=2 \\(x-2)(x^2+ax+b)=x^3+ax^2+bx-2x^2-2ax-2b=\\=x^3+x^2(a-2)+x(b-2a)-2b \\x^3-2x^2-9x+18=x^3+x^2(a-2)+x(b-2a)-2b \\ \left \{ {{a-2=-2} \atop {b-2a=-9}} \right. \Rightarrow \left \{ {{a=0} \atop {b=-9}} \right. \\(x-2)(x^2-9)=0 \\x^2-9=0 \\x^2=9 \\x_2=3 \\x_3=-3](https://tex.z-dn.net/?f=2%29x%5E3-2x%5E2-9x%2B18%3D0+%5C%5CP%282%29%3D8-8-18%2B18%3D0%5CRightarrow+x_1%3D2+%5C%5C%28x-2%29%28x%5E2%2Bax%2Bb%29%3Dx%5E3%2Bax%5E2%2Bbx-2x%5E2-2ax-2b%3D%5C%5C%3Dx%5E3%2Bx%5E2%28a-2%29%2Bx%28b-2a%29-2b+%5C%5Cx%5E3-2x%5E2-9x%2B18%3Dx%5E3%2Bx%5E2%28a-2%29%2Bx%28b-2a%29-2b+%5C%5C+%5Cleft+%5C%7B+%7B%7Ba-2%3D-2%7D+%5Catop+%7Bb-2a%3D-9%7D%7D+%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Ba%3D0%7D+%5Catop+%7Bb%3D-9%7D%7D+%5Cright.+%5C%5C%28x-2%29%28x%5E2-9%29%3D0+%5C%5Cx%5E2-9%3D0+%5C%5Cx%5E2%3D9+%5C%5Cx_2%3D3+%5C%5Cx_3%3D-3)
Ответ: x1=2; x2=3; x3=-3
Если что непонятно - спрашивай
=-<u> √3 </u> = -0,5√3
2
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6sin²x-3sinxcosx-cos²х=1
6sin²x-3sinxcosx-cos²х=sin²x+cos²х
6sin²x-3sinxcosx-cos²х-sin²x-cos²х=0
5sin²x-3sinxcosx-2cos²х=0 I :cos²х
5tg²x-3tgx-2=0
Введем новую переменную:
tgx=a
5a²-3a-2=0
D=9+40=49=7²
a1=(3+7)/10=1
a2=(3-7)/10=-2/5
Возвращаемся к замене:
tgx=1 tgx=-2/5
x=π/4+πn, n∈z x=-arctg2/5 + πn, n∈z