Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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
=> 4.(23+n)=3(40+n)
92+4n=120+3n
92-120=3n-4n
-28=-1n
=>n=28
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
suy ra (23+n).4=(40+n).3
suy ra 92+4n=120+3n
suy ra 4n-3n=120-92
suy ra n=28
\(\frac{23+n}{40+n}=\frac{3}{4}\)
\(\Leftrightarrow4\left(23+n\right)=3\left(40+n\right)\)
\(\Leftrightarrow92+4n=120+3n\)
\(\Rightarrow4n-3n=120-92\)
\(n=28\)
vậy n=28
nhớ kb và k nha