\(\dfrac{x+1}{3}=\dfrac{-3}{-9}\)

=>\(\dfrac{x+1}{3}=\dfrac{1}{3}\)

=>x+1=1

=>x=0

\(\dfrac{x+1}{3}=\dfrac{-3}{-9}\)

=>\(\dfrac{x+1}{3}=\dfrac{1}{3}\)

=>x+1=1

=>x=0

-3/10 -2/5

= -3/10 -4/10

= -7/10

\(\dfrac{-3}{10}-\dfrac{2}{5}=\dfrac{-7}{10}\)

50% + 1,5 : (1-1/3)

= 50% + 1,5 : (1-1/3)

= 50% + 1,5 : 2/3

= 1/2 + 15/10 : 2/3

= 1/2 + 45/20

= 10/20 + 45/20

= 55/20 = 2,75

\(B=\left(3-\dfrac{1}{5}\right)\left(3-\dfrac{2}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)

\(=\left(3-\dfrac{15}{5}\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)\)

\(=\left(3-3\right)\left(3-\dfrac{1}{5}\right)\cdot...\cdot\left(3-\dfrac{29}{5}\right)=0\)

31/42?

-3/20

a) Ta có:

3n - 5 = 3n + 12 - 17 = 3(n + 4) - 17

Để A nguyên thì 17 ⋮ (n + 4)

⇒ n + 4 ∈ Ư(17) = {-17; -1; 1; 17}

⇒ n ∈ {-21; -5; -3; 13}

b) Ta có:

n + 1 = n + 2 - 1

Để B nguyên thì 1 ⋮ (n + 2)

⇒ n + 2 ∈ Ư(1) = {-1; 1}

⇒ n ∈ {-3; -1}

c) 10n = 10n - 4 + 4= 2(5n - 2) + 4

Để C nguyên thì 4 ⋮ (5n - 2)

⇒ 5n - 2 ∈ Ư(4) = {-4; -2; -1; 1; 2; 4}

⇒ 5n ∈ {-2; 0; 1; 3; 4; 6}

⇒ n ∈ {-2/5; 0; 1/5; 3/5; 4/5; 6/5}

Mà n ∈ Z

⇒ n = 0

d) 6n - 1 = 6n + 4 - 5 = 2(3n + 2) - 5

Để D nguyên thì 5 ⋮ (3n + 2)

⇒ 3n + 2 ∈ Ư{5} = {-5; -1; 1; 5}

⇒ 3n ∈ {-7; -3; -1; 3}

⇒ n ∈ {-6/3; -1; -1/3; 1}

Mà n ∈ Z

⇒ n ∈ {-1; 1}