Gọi \(ƯCLN\left(5a+2b;7a+3b\right)=d\) \(\left(d\in N\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+2b⋮d\\7a+3b⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15a+6b⋮d\\14a+6b⋮d\end{matrix}\right.\)

\(\Leftrightarrow a⋮d\)

Mà \(5a+2b⋮d\) \(\Leftrightarrow b⋮d\)

\(\Leftrightarrow d⋮a,b\Leftrightarrow d⋮d'\left(1\right)\)

Gọi \(d'=ƯCLN\left(a,b\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a⋮d'\\b⋮d'\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+2b⋮d'\\7a+3b⋮d'\end{matrix}\right.\)

\(\Leftrightarrow d'⋮d\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

\(\text{Gọi:}d=UCLN\left(5a+2b,7a+3b\right)\Rightarrow\hept{\begin{cases}15a+6b⋮d\\14a+6b⋮d\end{cases}}\Rightarrow a⋮d;2a+b⋮d\Rightarrow b⋮d\)

do đó: \(UCLN\left(a,b\right)\ge UCLN\left(5a+2b,7a+3b\right);\text{mặt khác:}Goi:d'=UCLN\left(a,b\right)\Rightarrow\hept{\begin{cases}5a+2b⋮d'\\7a+3b⋮d'\end{cases}}\)

do đó:\(UCLN\left(5a+2b,7a+3b\right)\ge UCLN\left(a,b\right)\text{ suy ra điều phải chứng minh}\)

MIK vẫn chư hiểu đoạnƯCLN(a,b)>ƯCLN (5A+2B,7A+3B)

chưa hiểu lắm

Đề có sai ko bn

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2a+3b}{2c+3d}=\dfrac{2bk+3b}{2dk+3d}=\dfrac{b}{d}\)

\(\dfrac{2a-3b}{2c-3d}=\dfrac{2bk-3b}{2dk-3d}=\dfrac{b}{d}\)

Do đó: \(\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)

Bạn vào câu hỏi tương tự nhé  Dương Diệu Linh