Lời giải:
$a-11b+3c\vdots 17$

$\Rightarrow 2(a-11b+3c)\vdots 17$

$\Rightarrow 2a-22b+6c\vdots 17$

$\Rightarrow 2a-5b+6c-17b\vdots 17$

$\Rightarrow 2a-5b+6c\vdots 17$ (đpcm)

nhân 2a-5b+6c với 9 rồi trừ đi a-11b+3c

Ta có \(a-11b+3c⋮17\Rightarrow2a-22b+6c⋮17\)

Ta có \(17b⋮17\)

Nên \(2a-22b+6c+17b=2a-5b+6c⋮17\left(dpcm\right)\)

1duocgoitienganhla

Nguyễn Ngọc Ánh Minh trả lời đúng quá

Ta có:\(\left(2a-5b+6c\right)+15\left(a-11b+3c\right)=17a-170b+51c⋮17\)

Mà \(15\left(a-11b+3c\right)⋮17\Rightarrow2a-5b+6c⋮17\left(đpcm\right)\)

Nếu \(a-11b+3c⋮17\Rightarrow2\left(a-11b+3c\right)⋮17\)

\(\Rightarrow2a-22b+6c⋮17\Rightarrow\left(2a-5b+6c\right)-17b⋮17\)

Vì\(17b⋮17\Rightarrow2a-5b+3c⋮17\)

Vì \(a-11b+3c\) chia hết cho 17 => \(2\left(a-11b+3c\right)\)chia hết cho 17 =>      \(2a-22b+6c\)

Ta có:     \(\left(2a-22b+6c\right)-\left(2a-5b+6c\right)=17b\)chia hết  cho 17

Mà 2a - 22b + 6c chia hết cho 17 nên => 2a - 5b + 6c chia hết cho 17

Vậy 2a - 5b + 6c chia hết cho 17.

\(a-11b+3c⋮17\)\(\Rightarrow2\left(a-11b+3c\right)⋮17\)\(\Rightarrow2a-22b+6c⋮17\)

mà \(17b⋮17\)\(\Rightarrow\left(2a-22b+6c\right)+17b⋮17\)

\(\Rightarrow2a-22b+6c+17b⋮17\)\(\Rightarrow2a-5b+6c⋮17\left(đpcm\right)\)

Ta có \(a-11b+3c⋮17\)

     => \(19\left(a-11b+3c\right)⋮17\)

     => \(19a-209b+57c⋮17\)

     =>  ( 17a - 204b + 51c ) + ( 2a - 5b + 6c ) \(⋮\)17

     => 2a - 5b + 6c \(⋮\)17 ( do 17a - 204b + 51c \(⋮\)17 )   ( đpcm )