Question

Fie a,b,c trei nr naturale astfel in cat: a/b=7/15 SI b/c=3/8 . a) verificati egalitatea (a+b)^2(la puterea a 2a )=a^2+2ab+b^2 .

Answer 1

a/b=7/15 deci 7b=15a 
b/c=3/8 deci 8b=3c
a=7b/15
c=8b/3
$\frac{8b}{3}=5*(b-\frac{7b}{15}) \\ \frac{8b}{3}=5*\frac{8b}{15} \\ \frac{8b}{3}=\frac{8b}{3}$
$a+b+c=124 \\ \frac{7b}{15}+b+\frac{8b}{3}=124 \\ \frac{7b}{15}+\frac{15b}{15}+\frac{40b}{15}=\frac{1860}{15} \\ 62b=1860 =\ \textgreater \ b=30 \\ a=7*30:15=14 \\ c=8*30:3=80$

$(a+b)^2=(a+b)(a+b) \\ a*a+ab+b*b+ab=a^2+2ab+b^2$

Answer 2

$a)~a^2+2ab+b^2=a^2+ab+ab+b^2=a(a+b)+b(a+b)= \\ =(a+b)(a+b)=(a+b)^2. \\ \\ b)~ \frac{a}{b}= \frac{7}{15} \Rightarrow b= \frac{15a}{7}. \\ \\ \frac{b}{c}= \frac{3}{8} \Rightarrow c= \frac{8b}{3}= \frac{8* \frac{15a}{7} }{3}= \frac{120a}{21} = \frac{40a}{7} . \\ \\ b-a= \frac{15a}{7}-a= \frac{15a}{7}- \frac{7a}{7}= \frac{8a}{7}. \\ \\ 5(b-a)=5* \frac{8a}{7}= \frac{40a}{7} =c.$

$c)~a+b+c=124 \Leftrightarrow \\ \\ \Leftrightarrow a+ \frac{15a}{7}+ \frac{40a}{7}=124 \Leftrightarrow \\ \\ \Leftrightarrow 7a+15a+40a=7*124 \Leftrightarrow \\ \\ \Leftrightarrow 62a=868 \Rightarrow a=14. \\ \\ b=\frac{15a}{7}= \frac{15*14}{7}= 30. \\ \\ c= \frac{40a}{7}= \frac{40*14}{7}=80. \\ \\ \underline{Solutie}: (a,b,c)=(14,30,80).$