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Tori163
@Tori163
September 2021
2
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Разложите на множители:
а)100а^4-1/9b^2
b)9x²-(x-1)²
в)x³+y^6
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hraky
A) (10a^2-1/(3b))(
10a^2+1/(3b))
б) (3x-x+1)(3x+x-1)=(2x+1)(4x-1)
в) (x+y^2)(x^2-xy^2+y^4)
1 votes
Thanks 3
litanalove
A) 10²a^4-1/3²b^4=
=(10a²-1/3b²)(10a²+1/3b²)
b) пусть (x-1)=a, тогда
9х²-a²=(3x-a)(3x+a)=
=(3x-x+1)(3x+x-1)=
=(2x-1)(4x+1)
в) x^3+y^6=(x+y²)(х²-xy²+y^4)
2 votes
Thanks 2
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Answers & Comments
б) (3x-x+1)(3x+x-1)=(2x+1)(4x-1)
в) (x+y^2)(x^2-xy^2+y^4)
=(10a²-1/3b²)(10a²+1/3b²)
b) пусть (x-1)=a, тогда
9х²-a²=(3x-a)(3x+a)=
=(3x-x+1)(3x+x-1)=
=(2x-1)(4x+1)
в) x^3+y^6=(x+y²)(х²-xy²+y^4)