Question

cmr nếu 3a+4b+5c chia hết cho11 thì 9a+b+4c cũng chia hết cho 11

Answer 1

3a + 4b + 5c \(⋮\) 11

\(\Rightarrow\)3.(3a + 4b + 5c) = 9a + 12b + 15c \(⋮\) 11

\(\Rightarrow\) (9a + 12b + 15c) - (11b + 11c) = 9a + b + 4c \(⋮\)11

Answer 2

Ta có: \(\left(3a+4b+5c\right)⋮11\)

\(\Rightarrow3\left(3a+4b+5c\right)⋮11\)(1)

Ta lại có: \(11\left(b+c\right)⋮11\forall b,c\)(2)

Từ (1) và (2) suy ra \(3\left(3a+4b+5c\right)-11\left(b+c\right)⋮11\)

hay \(9a+b+4c⋮11\)(đpcm)

Answer 3

☘ Ta có 3a + 4b + 5c ⋮ 11

⇒ 3(3a + 4b + 5c) ⋮ 11

3(3a + 4b + 5c)

= (9a + 12b + 15c) ⋮ 11

☘ 11b + 11c ⋮ 11

⇒ (9a + 12b + 15c) - (11b + 11c)

= 9a + b + 4c

Mà (9a + 12b + 15c) ⋮ 11

11b + 11c ⋮ 11

Nên (9a + 12b + 15c) - (11b + 11c) = 9a + b + 4c ⋮ 11

➤ 9a + b + 4c ⋮ 11