\(\Rightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(78+32-\left\{90-1^{100}+\left(5^2+15\right)\right\}\)

\(=110-90+1^{100}-\left(25+15\right)\)

=20+1-40

=21-40=-19

tk

-19

???

a/

Gọi d là ước chung của 2n+1 và 3n+1 nên

\(2n+1⋮d\Rightarrow3\left(2n+1\right)=6n+3⋮d\)

\(3n+1⋮d\Rightarrow2\left(3n+1\right)=6n+2⋮d\)

\(\Rightarrow6n+3-\left(6n+2\right)=1⋮d\Rightarrow d=1\)

Điều đó chứng tỏ rằng 2n+1 và 3n+1 là 2 số nguyên tố sánh đôi

Các câu b;c;d làm tương tự

\(\dfrac{-4}{9}\cdot x=\dfrac{-2}{7}:\dfrac{4}{21}\\ -\dfrac{4}{9}\cdot x=\dfrac{-2}{7}\cdot\dfrac{21}{4}\\ -\dfrac{4}{9}\cdot x=\dfrac{-3}{2}\\ x=-\dfrac{3}{2}:\dfrac{-4}{9}\\ x=\dfrac{3}{2}\cdot\dfrac{9}{4}\\ x=\dfrac{27}{8}\)

Vậy: ... 

\(a.BC\left(3\cdot5^2;5^2\cdot7\right)\\ =B\left(3\cdot5^2\cdot7\right)\\ =B\left(525\right)=\left\{0;525;1050;...\right\}\\ b.ƯC\left(2^2\cdot3\cdot5;3^2\cdot7;3\cdot5\cdot11\right)\\ =Ư\left(3\right)=\left\{1;3\right\}\)

tìm cả bội chung và ức chung của cả hai 

\(\dfrac{3}{7}\cdot x=-\dfrac{1}{3}:\dfrac{2}{9}\\ \dfrac{3}{7}\cdot x=-\dfrac{1}{3}\cdot\dfrac{9}{2}\\ \dfrac{3}{7}\cdot x=-\dfrac{3}{2}\\ x=-\dfrac{3}{2}:\dfrac{3}{7}\\ x=-\dfrac{7}{2}\)

Vậy: ... 

`3/7x = -1/3 : 2/9`

`=> 3/7x = -1/3 . 9/2`

`=> 3/7 x = -3/2`

`=> x = -3/2 . 7/3`

`=> x = -7/2`

\(\dfrac{5}{21}\cdot\dfrac{-3}{22}-\left(-\dfrac{4}{21}\right)+\dfrac{5}{21}\cdot\dfrac{-19}{22}\\ =\dfrac{5}{21}\cdot\left(-\dfrac{3}{22}-\dfrac{19}{22}\right)+\dfrac{4}{21}\\ =\dfrac{5}{21}\cdot\dfrac{-22}{22}+\dfrac{4}{21}\\ =\dfrac{-5}{21}+\dfrac{4}{21}\\ =\dfrac{-1}{21}\)

`5/13 + 2/7 + 8/13 - 3/14`

`= (5/13 + 8/13) + (2/7 - 3/14)`

`= 13/13 + 1/14`

`= 1 + 1/14 `

`= 15/14`

\(\left(27^3\right)^4=27^{3\cdot4}=27^{12}\)