Question

Докажите тождество:1) (3m-7)*0,6-0,8(4m-5)-(-1,7-1,4m)=1,5;2) 7а(3b+4c)-3а (b+1/3с)=9а(2b+3с)??

Answer 1

1) 
$\displaystyle (3m-7)*0.6-0.8(4m-5)-(-1.7-1.4m)=1.5\\1.8m-4.2-3.2m+4+1.7+1.4m=1.5\\(1.8-3.2+1.4)m+4-4.2+1.7=1.5\\0m+1.5=1.5\\1.5=1.5$

2) 
$\displaystyle 7a(3b+4c)-3a(b+ \frac{1}{3}c)=9a(2b+3c)\\21ab+28ac-3ab-ac=9a(2b+3c)\\18ab+28ac=9a(2b+3c)\\9a(2b+3c)=9a(2b+3c)$