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tatyanapopova2
@tatyanapopova2
June 2022
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(х-4)в кубе-(х-7)в квадрате(х+2)-27=0
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(a-b)²=a²-2ab+b²
(a-b)³=a³-3a²b+3ab²-b³
(x-4)³-(x-7)²(x+2)-27=0
1) (x-4)³=x³-3x²*4+3x*4²-4³=x³-12x²+48x-64
2) (x-7)²=x²-14x+49
3) (x²-14x+49)(x+2)=x³-14x²+49x+2x²-28x+98=x³-12x²+21x+98
4) x³-12x²+48x-64-(x³-12x²+21x+98)=x³-12x²+48x-64-x³+12x²-21-98=27x-189=
=27(x-7)
27(x-7)=0
x-7=0
x=7
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Answers & Comments
(a-b)³=a³-3a²b+3ab²-b³
(x-4)³-(x-7)²(x+2)-27=0
1) (x-4)³=x³-3x²*4+3x*4²-4³=x³-12x²+48x-64
2) (x-7)²=x²-14x+49
3) (x²-14x+49)(x+2)=x³-14x²+49x+2x²-28x+98=x³-12x²+21x+98
4) x³-12x²+48x-64-(x³-12x²+21x+98)=x³-12x²+48x-64-x³+12x²-21-98=27x-189=
=27(x-7)
27(x-7)=0
x-7=0
x=7