\(a+b+c=14\)
\(7a+7b+7c=14.7=98\)
Ta có: \(9a+8a+6c-\left(7a+7b+7c\right)=9a+8b+6c-7a-7b-7c=2a+2b-c\)
Do đó: \(2a+2b-c=101-98=3\)
\(9a+8b+6c=7a+2a+7b+b+7c-c\)
\(=7a+7b+7c+\left(2a+b-c\right)=7.\left(a+b+c\right)+\left(2b+b-c\right)=101\)
\(2a+b-c=101-7.\left(a+b+c\right)\)
mk làm thiếu nha sorry :>
\(101-7.\left(a+b+c\right)=101-7.14=3\)