решите неравенство1) (5-х)(х+5)-х(х+3)<6
2)(x-4)²-x²≥0
3)[tex](x-1)^{3}- (x+1)^{3} \leq x-6 x^{2} [/tex]
4)[tex] (3+x)^{2}- (3-x)^{2}\ \textgreater \ 16x-21[/tex]
2)(x-4)²-x²≥0
3)[tex](x-1)^{3}- (x+1)^{3} \leq x-6 x^{2} [/tex]
4)[tex] (3+x)^{2}- (3-x)^{2}\ \textgreater \ 16x-21[/tex]
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(5-х)(х+5)-х(х+3)<6
25-х²-х²-3х<6
25-3х<6
-3x<-19
x>19|3
x>6 1|3
Отвт:(6 1/3;∞)
2)
(х-4)²-х²≥0
(х-4-х)(х-4+х)≥0
-4(2х-4)≥0
2х-4≤0
2х≤4
х≤2
Ответ:(-∞;2]
3)
(х-1)³-(х+1)³≤х-6х²
(х-1-х-1)((х-1)²+(х-1)(х+1)+(х+1)²)≤х-6х²
-2(х²-2х+1+х²-1+х²+2х+1)≤х-6х²
-2×(3х²+1)≤х-6х²
-6х²-2-х-6х²≤0
-2-х≤0
-х≤2
х≥-2
Ответ:[-2;∞)
4)
(3+х)²-(3-х)²>16x-21
(3+x-3+x)(3+x+3-x)>16x-21
2x×6>16x-21
12x-16x>-21
-4x>-21
x<21/4
х<5 1/4
Ответ:(-∞;5 1/4)