Предмет: Алгебра,
автор: lelyp
разложите на множители y^2-2y-24 и 64y^4+1
Ответы
Автор ответа:
0
1)
![y^{2} -2y-24= y^{2} -6y+4y-24=y(y-6)+4(y-6)=\
(y+4)(y-6)](https://tex.z-dn.net/?f=+y%5E%7B2%7D+-2y-24%3D+y%5E%7B2%7D+-6y%2B4y-24%3Dy%28y-6%29%2B4%28y-6%29%3D%5C%0A%28y%2B4%29%28y-6%29 "y^{2} -2y-24= y^{2} -6y+4y-24=y(y-6)+4(y-6)=\
(y+4)(y-6)")
2)
![64 y^{4}+1=64y ^{4} +16 y^{2} -16 y^{2}+1=(8y ^{2} +1) ^{2} -16 y^{2} =\
(8y ^{2}+1) ^{2} -4 ^{2} y^{2} =(8y ^{2} +1-4y)(8 y^{2}+1+4y)](https://tex.z-dn.net/?f=64+y%5E%7B4%7D%2B1%3D64y+%5E%7B4%7D++%2B16+y%5E%7B2%7D+-16+y%5E%7B2%7D%2B1%3D%288y+%5E%7B2%7D++%2B1%29+%5E%7B2%7D+-16+y%5E%7B2%7D+%3D%5C%0A%288y+%5E%7B2%7D%2B1%29+%5E%7B2%7D++-4+%5E%7B2%7D+y%5E%7B2%7D+%3D%288y+%5E%7B2%7D++%2B1-4y%29%288+y%5E%7B2%7D%2B1%2B4y%29+ "64 y^{4}+1=64y ^{4} +16 y^{2} -16 y^{2}+1=(8y ^{2} +1) ^{2} -16 y^{2} =\
(8y ^{2}+1) ^{2} -4 ^{2} y^{2} =(8y ^{2} +1-4y)(8 y^{2}+1+4y)")
2)
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