Задание 477

Решение:

• \({\largeа)}\ x^4-3x^2+2=x^4-x^2-2x^2+2=x^2(x^2-1)-2(x^2-1)=(x^2-1)(x^2-2)=(x+1)(x-1)(x^2-2)\)

• \({\largeб)}\ b^2c^2-4bc-b^2-c^2+1=b^2c^2-2bc+1-b^2-2bc-c^2=b^2c^2-2bc+1-(b^2+2bc+c^2)=(bc-1)^2-(b+c)^2=((bc-1)+(b+c))((bc-1)-(b+c))=(bc-1+b+c)(bc-1-b-c)\)

• \({\largeв)}\ y^2-10y+25-4x^2=(y-5)^2-(2x)^2=(y-5+2x)(y-5-2x)\)

• \({\largeг)}\ (a+b)^3-a^3-b^3=(a+b)^3-(a^3+b^3)=(a+b)(a+b)^2-(a+b)(a^2-ab+b^2)=(a+b)(a^2+2ab+b^2-(a^2-ab+b^2))=(a+b)(\)\(a^2\) \(+\) \(2ab\) \(+\) \(b^2\) \(-\) \(a^2\) \(+\) \(ab\) \(-\) \(b^2\)\()=3ab(a+b)\)

• \({\largeд)}\ x^{16}-y^{16}=(x^8)^2-(y^8)^2=(x^8+y^8)(x^8-y^8)=(x^8+y^8)((x^4)^2-(y^4)^2)=(x^8+y^8)(x^4+y^4)(x^4-y^4)=(x^8+y^8)(x^4+y^4)((x^2)^2-(y^2)^2)=(x^8+y^8)(x^4+y^4)(x^2+y^2)(x^2-y^2)=(x^8+y^8)(x^4+y^4)(x^2+y^2)(x+y)(x-y)\)

• \({\largeе)}\ x^4-3x^2+1=x^4-2x^2+1-x^2=(x^2-1)^2-x^2=(x^2-1+x)(x^2-1-x)\)

• \({\largeж)}\ x^4-8x^2+4=x^4-4x^2+4-4x^2=(x^2-2)^2-(2x)^2=(x^2-2+2x)(x^2-2-2x)\)

• \({\largeз)}\ x^4-7x^2+1=x^4+2x^2+1-9x^2=(x^2+1)^2-(3x)^2=(x^2+1+3x)(x^2+1-3x)\)

• \({\largeи)}\ x^4+12x^2+64=x^4+16x^2+64-4x^2=(x^2+8)^2-(2x)^2=(x^2+8+2x)(x^2+8-2x)\)

• \({\largeк)}\ x^4+x^2-2=x^4-2x^2+1+3x^2-3=(x^2-1)^2+3(x^2-1)=(x^2-1)(x^2-1+3)=(x+1)(x-1)(x^2+2)\)