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KSYLYA
@KSYLYA
December 2021
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4*(5-x)^2-3*(x-5)разложить. на множитель
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slavaekb
4*(25-10x+x^2)-3x+15
100-40x+4x^2-3x+15
4x^2-43x+115
D=43^2-4*4*115=1849-1840=9
X1,2=(43+-3)/8=
x=23/4
x2=5
разложим на множители
(x-5)*(4x-23)
как то так
0 votes
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Answers & Comments
100-40x+4x^2-3x+15
4x^2-43x+115
D=43^2-4*4*115=1849-1840=9
X1,2=(43+-3)/8=
x=23/4
x2=5
разложим на множители
(x-5)*(4x-23)
как то так