Решите неравенство: 1) (х-4)(х+3)-5 < х(х+16) 2) х²(6-х)+х³ ≥ 2х(3х+4)-24 3) х³(3-2х)+х²(х+2х²) > 4х(х²-3)+12. Created by vika050901. algebra-ru

1)x^2-x-17 < x^2+16 xx^2-x-17-x^2-16 x < 0-17 x-17 < 0-17 x < 17x > -12)6x^2-x^3+x^36x^2+8x-246x^2-x^3+x^3-6x^2-8x+24-8x+24-8xx3)3x^3-2x^4+x^3+2x^4>4x^3-12x+123x^3+x^3-4x^3+12x-12>012x-12>012x>12x>1

Решите неравенство: 1) (х-4)(х+3)-5 < х(х+16) 2) х²(6-х)+х³ ≥ 2х(3х+4)-24 3)...

vika050901 @vika050901

August 2022 1 3 Report

Решите неравенство:
1) (х-4)(х+3)-5 < х(х+16)
2) х²(6-х)+х³ ≥ 2х(3х+4)-24
3) х³(3-2х)+х²(х+2х²) > 4х(х²-3)+12

Answers & Comments

ekaterinkamiss 1)x^2-x-17 < x^2+16 x
x^2-x-17-x^2-16 x < 0
-17 x-17 < 0
-17 x < 17
x > -1
2)6x^2-x^3+x^3 $\geq$ 6x^2+8x-24
6x^2-x^3+x^3-6x^2-8x+24 $\geq 0$
-8x+24 $\geq 0$
-8x $\geq -24$
x $\leq 3$
3)3x^3-2x^4+x^3+2x^4>4x^3-12x+12
3x^3+x^3-4x^3+12x-12>0
12x-12>0
12x>12
x>1

4 votes Thanks 5

vika050901 Спасибо огромное)

ekaterinkamiss незачто))))))))

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