1
(ab+ac)²/(ab²-ac²)=a²(b+c)/a(b²-c²)=a(b+c)/(b-c)(b+c)=a/(b-c)
2
1)1/c-1/(10-c)=(10-c-c)/c(10-c)=(10-2c)/c(10-c)=2(5-c)/c(10-c)=2(c-5)/c(c-10)
2)25/(c-5) -(c-5)=(25-c²+10c-25)/(c-5)=(10c-c²)/(c-5)=c(10-c)/(c-5)
3)2(c-5)/c(c-10) * c(10-c)/(c-5)=-2
Ответ: x∈(1;+∞).
Объяснение:
(5x-3)/4-(3-x)/5>(2-x)/10 |×20
5*(5x-3)-4*(3-x)>2*(2-x)
25x-15-12+4x>4-2x
31x>31 |÷31
x>1.
Согласно условиям:
а) х+3у<11
б) y/x<1