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azizovafilis7
@azizovafilis7
April 2022
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срочно надо,пожалуйста помогите
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zinaidazina
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azizovafilis7
Спасибо огромное❤
zinaidazina
Удачи!
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Answers & Comments
1)![0,027x^3+1=(0,3x)^3+1=(0,3x+1)(0,09x^{2} -0,3x+1) 0,027x^3+1=(0,3x)^3+1=(0,3x+1)(0,09x^{2} -0,3x+1)](https://tex.z-dn.net/?f=0%2C027x%5E3%2B1%3D%280%2C3x%29%5E3%2B1%3D%280%2C3x%2B1%29%280%2C09x%5E%7B2%7D%20-0%2C3x%2B1%29)
2)![x^{3}+8y^3z^9=(x+2yz^3)(x^{2} -2xyz^3+4y^2z^6) x^{3}+8y^3z^9=(x+2yz^3)(x^{2} -2xyz^3+4y^2z^6)](https://tex.z-dn.net/?f=x%5E%7B3%7D%2B8y%5E3z%5E9%3D%28x%2B2yz%5E3%29%28x%5E%7B2%7D%20-2xyz%5E3%2B4y%5E2z%5E6%29)
3)![8x^3+y^6=(2x+y^2)(4x^{2} -2xy^2+y^4) 8x^3+y^6=(2x+y^2)(4x^{2} -2xy^2+y^4)](https://tex.z-dn.net/?f=8x%5E3%2By%5E6%3D%282x%2By%5E2%29%284x%5E%7B2%7D%20-2xy%5E2%2By%5E4%29)
4)![x^9+\frac{1}{64}=(x^3+\frac{1}{4})(x^{6} - \frac{1}{4}x^3+\frac{1}{16}) x^9+\frac{1}{64}=(x^3+\frac{1}{4})(x^{6} - \frac{1}{4}x^3+\frac{1}{16})](https://tex.z-dn.net/?f=x%5E9%2B%5Cfrac%7B1%7D%7B64%7D%3D%28x%5E3%2B%5Cfrac%7B1%7D%7B4%7D%29%28x%5E%7B6%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7Dx%5E3%2B%5Cfrac%7B1%7D%7B16%7D%29)