Question

Answer 1

Lời giải:
Ta có: $\frac{31-n}{43-n}=\frac{3}{5}$

$\frac{5\times (31-n)}{5\times (43-n)}=\frac{3\times (43-n)}{5\times (43-n)}$

$5\times (31-n)=3\times (43-n)$

$5\times 31-5\times n=3\times 43-3\times n$

$155-5\times n=129-3\times n$

$155-5\times n-(129-3\times n)=0$

$155-5\times n-129+3\times n=0$

$(155-129)-(5\times n-3\times n)=0$

$26-2\times n=0$

$2\times n=26$

$n=13$

Answer 2

13

Answer 3

 

Theo đề, ta có: 

\(\dfrac{31-n}{43-n}=\dfrac{3}{5}\)

\(\Leftrightarrow155-5n=129-3n\)

\(\Leftrightarrow-2n=-26\)

hay n=13