Question

Решите неравенство 4-5х>2-3(х-1) и 3-2(х+1)<5+х

Разложите на множители $a^{2} -9b ^{2} +18bc-9 c^{2}$ 

Answer 1

4-5х>2-3(х-1)
4-5х>2-3х+3
-2х<1
х>-1/2
х€(-оо;-1/2)

3-2(х+1)<5+х
3-2х-2<5+х
-х<4
х>-4
х€(-оо;-4)

а²-9b²+18bc-9c²=(a-3b)(a+3b)+9c(2b-c)

Answer 2

4-5х>2-3(х-1)
4-5х>2-3х+3
-2x>1
x<-0.5

3-2(х+1)<5+х
3-2х-2<5+х
-3x<4
x>-4/3
$a^2-9b^2+18bc-9c^2= a^2-9(b^2-2bc+c^2)=a^2-9(b-c)^2=(a-3b+3c)(a+3b-3c)$