Ta có  \(\frac{23+n}{40+n}=\frac{3}{4}\) 

\(\Rightarrow\left(23+n\right)\times4=\left(40+n\right)\times3\)

\(\Rightarrow92+4n=120+3n\) 

\(\Rightarrow4n-3n=120-92\) 

\(\Rightarrow1n=28\)

\(\Rightarrow x=28\in Z\) 

Vậy x = 28

Thank you, cảm ơn bạn rất nhiều!!!!!!!!!!!!!!!!!

23+x/40+x=3/4

=>x/40+x=3/4-23=-89/4

=>x/40+40x/40=-89/4

=>(x+40x)/40=-89/4

=>50x/40=-89/4

=>50x=-89/4.40=-890

=>x=-17,8

k cho mk nha pn!!!!!

\(\frac{23+n}{40+n}=\frac{3}{4}\)

\(\Rightarrow4\left(23+n\right)=3\left(40+n\right)\)

\(\Rightarrow92+4n=120+3n\)

\(\Rightarrow\left(92+4n\right)-\left(92+3n\right)=\left(120+3n\right)-\left(92+3n\right)\)

\(\Rightarrow n=28\)

Ta có: \(\frac{23+n}{40+n}\)=\(\frac{3}{4}\)\(\Rightarrow\)4(23+ n)= 3(40+ n)\(\Rightarrow\)92+ 4n= 120+ 3n\(\Rightarrow\)4n- 3n= 120- 92\(\Rightarrow\)(4- 3)n= 82\(\Rightarrow\)n= 82.

Vậy n= 82 để  \(\frac{23+n}{40+n}\)=\(\frac{3}{4}\).

\(\frac{23+x}{40+x}=\frac{3}{4}\)

\(\Leftrightarrow4\left(23+x\right)=3\left(40+x\right)\)

\(\Leftrightarrow92+4x=120+3x\)

\(\Leftrightarrow4x-3x=120-92\)

\(\Leftrightarrow x=28\)

\(\frac{23+n}{40+n}=\frac{3}{4}\Leftrightarrow4\left(23+n\right)=3\left(40+n\right)\)

\(\Leftrightarrow92+4n=120+3n\)

\(\Leftrightarrow92-120=-4n+3n\)

\(\Leftrightarrow-28=-n\Leftrightarrow n=28\)

duyệt nha bn

tốt quá tôi khâm phục bạn

=> 4.(23+n)=3(40+n)

92+4n=120+3n

92-120=3n-4n

-28=-1n

=>n=28

tuyệt bạn giỏi đấy

lam phan b nhe

23+n/40+n = 3/4

=>(23+n).4=(40+n).3

=>4n+92=3n+120

=>4n-3n=120-92

=>n=28

hỏi j vậy bn ?