1) a(3a2 – 4b)+b(4a-a2) 2)c(c4-c3+c2)-(c4-c3+c2) 3)19-2(5x+4)=10

Question

1) a(3a2 – 4b)+b(4a-a2) 2)c(c4-c3+c2)-(c4-c3+c2) 3)19-2(5x+4)=10

Answer 1

1) a(3a²-4b)+b(4a-a²)
3а³-4ab+4ab-a²b
3a³-a²b
2) c(c⁴-c³+c²)-(c⁴-c³+c²)
(c⁴-c³+c²)×(c-1)
c²(c²-c+1)(c-1)
3) 19-2(5x+4)=10
19-10x-8=10
11-10x=10
-10x=10-11
-10x=-1|÷(-10)
x=1/10

Answer 2

1)3a³-4ab+4ab-a²b=3a³-a²b

2)c5-c⁴+c³-c⁴+c³-c²=c5-2c⁴+2c³-c²

3)19-10x-8=10

8-10x=10

-10x=10-8

-10x=2

x=2/-10

x=-1/5

(2/-10;-1/5)-дробь