Ta có:
\(4a+5b+7c ⋮11\)
\(\Rightarrow\)\(16a+20b+28c⋮11\)
\(\Rightarrow\)\(5a+11a+9b+11b+22c+6c⋮11\)
\(\Rightarrow\)\(5a+9b+6c⋮11 \)\(\left(do\left\{{}\begin{matrix}11a⋮11\\11b⋮11\\22c⋮11\end{matrix}\right.\right)\)
Vậy: Nếu \(4a+5b+7c ⋮11\) thì \(5a+9b+6c ⋮11\)\(\left(ĐPCM\right)\)
4a + 5b + 7c \(⋮\) 11
=> 4.( 4a + 5b + 7c ) \(⋮\) 11
=> 16a + 20b + 28c \(⋮\) 11
=> ( 5a + 11a ) + ( 9b + 11b ) + ( 6c + 22c ) \(⋮\) 11
=> 5a + 11a + 9b + 11b + 6c + 22c \(⋮\) 11
=> ( 5a + 6b + 6c ) + ( 11a + 11b + 22c ) \(⋮\) 11
11a + 11b + 22c \(⋮\) 11 vì 11a \(⋮\) 11 ; 11b \(⋮\) 11 ; 22c \(⋮\) 11
=> 5a + 6b + 6c \(⋮\) 11 .
