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buholc2011
@buholc2011
August 2021
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3(2-x)^2-(2x^2+x-5)(x^2-2)+(x^2+4)(4-x^2)
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irreversibility
Verified answer
3(2-x)²-(2x²+x-5)(x²-2)+(x²+4)(4-x²)=3(4-4x+x²)-(2x⁴-4х²+х³-2х-5х²+10)+4x²-x⁴+16-4x²=12-12x+3x²-(2x⁴-9x²+x³-2x+10)-x⁴+16=12-12x+3x²-2x⁴+9x²-x³+2x-10-x⁴+16=18-10x+12x²-3x⁴-x³
13 votes
Thanks 23
buholc2011
Спасибо, а можешь преобразовать алгебраическое выражение в многочлен стандартного вида
irreversibility
какое выражение?
buholc2011
вот это 3(2-x)^2-(2x^2+x-5)(x^2-2)+(x^2+4)(4-x^2)
buholc2011
ау
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Verified answer
3(2-x)²-(2x²+x-5)(x²-2)+(x²+4)(4-x²)=3(4-4x+x²)-(2x⁴-4х²+х³-2х-5х²+10)+4x²-x⁴+16-4x²=12-12x+3x²-(2x⁴-9x²+x³-2x+10)-x⁴+16=12-12x+3x²-2x⁴+9x²-x³+2x-10-x⁴+16=18-10x+12x²-3x⁴-x³