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roman00
@roman00
July 2022
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Сюррикен
ОДЗ: x^2+3x-4 ≠ 0
D=9+16= 25; √25=±5;
x(1,2)= -3±5/2 ≠ -4; 1;
(-∞;-4) и(1; +∞) положительные решения.
x^2+5x-9 -(x^2+3x-4)≥0
2x-5≥0
x≥5/2
=> [5/2;+∞].
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Answers & Comments
D=9+16= 25; √25=±5;
x(1,2)= -3±5/2 ≠ -4; 1;
(-∞;-4) и(1; +∞) положительные решения.
x^2+5x-9 -(x^2+3x-4)≥0
2x-5≥0
x≥5/2
=> [5/2;+∞].