а) 6x+3x^3y-12xy^2= 3x(2+x^2y-12y^2)
b) 54a-103a+102a= a(54-103+102)
sinα+cosα-sin(α-π/6)+cos(α-π/6)=√6cos(α-π/12);
sinα-sin(α-π/6)+cosα+cos(α-π/6)=
2sin((a-a+π/6)/2)cos((a+a-π/6)/2)+2cos((a+a-π/6)/2)cos((a-a+π/6)/2)=
2sin(π/12)cos(a-π/12)+2cos(a-π/12)cos(π/12)=
2*(1/2)cos(a-π/12)+2(√3/2)cos(a-π/12)=
cos(a-π/12)+√3cos(a-π/12)=(1+√3)cos(a-π/12);
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(1+√3)cos(a-π/12)≠√6cos(α-π/12);
1+√3≠√6;
Может неправильно переписали?
Ответ:
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Объяснение: