Question

Afla numarul necunoscut
4532÷[72×5+(9348÷4-n)]=2

(565+35)÷30÷4=(118+294×3) ÷(25×m )

(n+78) ÷5+312=332

(234÷n)+340÷10=60

(a×20+a×10)=10

Answer 1

$4532:[72\cdot 5+(9348 : 4-n)]=2 \\ 4532:[72 \cdot 5+(2337-n)]=2 \\ 72 \cdot 5+(2337-n) =4532:2 \\ 72 \cdot 5+(2337-n)=2266 \\ 360+2337-n=2266 \\ -n=2266-360-2337 \\ -n=-431 \\ n=431$

$(565+35):30:4=(118+294\cdot 3) :(25\cdot n ) \\ 600:30:4=(118+882):25n \\ 20:4=1000:25n \\ 5=40n \\ n=40:5 \\ n=8$

[tex](n+78) :5+312=332 \\ (n+78):5=332-312 \\ (n+78):5=20 \\ n+78=20 \cdot 5 \\ n+78=100 \\ n=100-78 \\ n=22[/tex]

$(234:n)+340:10=60 \\ (234:n)+34=60 \\ 234:n=60-34 \\ 234:n=26 \\ n=234:26 \\ n=9$

$\displaystyle (a\cdot20+a\cdot10)=10 \\ 20a+10a=10 \\ 30a=10 \\ a= \frac{10}{30} \Rightarrow a= \frac{1}{3}$