Ответ:
Доказано
Объяснение:
1) a³b³c²+a²b⁴c²+a²b³c³=0
a²b³c²×(a+b+c)=0, при a+b+c = 0
(a²b³c²)×0 = 0
0 = 0
2)a⁶b⁴-2a⁵b⁵-a⁴b⁶=a⁴b⁴
a⁴b⁴×(a²-2ab - b²)=a⁴b⁴ |:(a⁴b⁴), а≠0 и b≠0
a²-2ab-b² = 1
a²- b² = 2ab + 1
1) 7z/4(x-y)
3) 34b/105(b+c)
1) 6y+x/2xy*(x - y)
2) 4b+c/2bc*(3c-1)
1) - a+2ab/b2-ab
3) 5b2-3a2/ab*(x-y)
1) 5x+3/2(x2-9)
3)2+2c2+5c/4c2-25
1)3m-8m3+2m2+27/(2m-9)*(9-4m2)
2) 23n+6n2-5/(3n+5)*(3n-5)2
1) 5+12z-2z2/(6-z)2
2)21+93m+2m2-28m2/21m2-m3-146m+336
3)-3a+15a2/(-3a+1)2*(1+3a)
4) не знаю
Х²-11х+24>0
D=121-4*24=25
x₁=(11+5)/2=8
x₂=(11-5)/2=3
Ответ. (-∞;3)U(8;+∞)
5x-4y=13
2x-y=4 /*4⇒8x-4y=16
отнимем
-3х=-3
х=1
2-у=4
у=-2
(1;-2)