Biểu thức chia hết cho 2 là

1831 - 675 vì 11 -5 = 6 nên 1831 - 675 ⋮  2

2014.2015 vì 2014 ⋮ cho 2 Nên 2014 .2015 ⋮  2 

1.2.3.4....10 + 2014 vì 1.2.3.....10 ⋮2 ; 2014⋮2 nên 1.2....10+2014 ⋮ 2

1.3.5.7.9.11-105 vì 5 nhân với số lẻ nào tận cùng cũng bằng 5 

mà 5-5 = 0 ; 0 : 2 nên 1.3.5.7.9.11-105 ⋮2

Biểu thức chia hết cho 5 là 

2640+ 1020 + 15 vì tận cùng là 0 và 5 nên 2640 + 1020 + 15 ⋮ 5 

1373 + 192 vì 3+2 = 5 ⋮ 5 nên 1373 + 192 ⋮ 5 

2014 .2015 vì 2015 ⋮ 5 nên 2014 .2015 ⋮ 5

1.3.5.7.9.11- 105 vì 1.3.5.7.9.11 có số5 mà 5⋮5 nên 1 .3. 5. 7 9. 11 ⋮ 5

105 ⋮ 5 nên 1.3.5.7.9.11-105 ⋮5

giúp mình ạ 

1 năm có 4 quý
Quý II gồm tháng 4; 5; 6

\(\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(2^4-4^2\right)\)

\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(16-16\right)\)

\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).0\)

\(=0\)

-------------------------------------------------

\(\left(7^{1997}-7^{1995}\right):\left(7^{1994}.7\right)\)

\(=\left[7^{1995}\left(7^2-1\right)\right]:7^{1995}\)

\(=7^{1995}.48:7^{1995}\)

\(=48\)

-------------------------------------------------

\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)

\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(6561-6561\right)\)

\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)

\(=0\)

-------------------------------------------------

\(\left(2^8+8^3\right):\left(2^5.2^3\right)\)

\(=\left[2^8+\left(2^3\right)^3\right]:2^8\)

\(=\left(2^8+2^9\right):2^8\)

\(=2^8.\left(1+2\right):2^8\)

\(=2^8.3:2^8\)

\(=3\)

sos

Ta có:

\(5=5\)

\(24=2^3.3\)

\(BCNN\left(5;24\right)=2^3.3.5=120\)

______________

Ta có:

\(17=17\)

\(27=3^3\)

\(BCNN\left(17;27\right)=17.3^3=459\)

______________

Ta có:

\(45=3^2.5\)

\(48=2^4.3\)

\(BCNN\left(45;48\right)=3^2.5.2=720\)

______________

Ta có:

\(8=2^3\)

\(1=1\)

\(12=2^2.3\)

\(BCNN\left(8;1;12\right)=2^3.1.3=24\)

\(BCNN\left(5,24\right)=120\)

\(BCNN\left(17,27\right)=459\)

\(BCNN\left(45,48\right)=720\)

\(BCNN\left(8,12,1\right)=24\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

`x + 12 = 25`

`\Rightarrow x = 25 - 12`

`\Rightarrow x = 13`

Vậy, `x = 13`

`b)`

`134 - x = 125 \div 5`

`\Rightarrow 134 - x = 25`

`\Rightarrow x = 134 - 25`

`\Rightarrow x=109`

Vậy, `x = 109`

`c)`

`2 \times x - 53 = 22`

`\Rightarrow 2x = 22 + 53`

`\Rightarrow 2x = 75`

`\Rightarrow x = 75 \div 2`

`\Rightarrow x =37,5`

Vậy, `x = 37,5`

`d)`

`7 + 3 \times x = 16`

`\Rightarrow 3x = 16 - 7`

`\Rightarrow 3x =9`

`\Rightarrow x = 9 \div 3`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`2,`

`a)`

`(x + 3) - 15 = 2`

`\Rightarrow x + 3 = 2 + 15`

`\Rightarrow x +3 = 17`

`\Rightarrow x = 17 - 3`

`\Rightarrow x = 14`

Vậy, `x = 14`

`b)`

`33 + (x - 1) = 56`

`\Rightarrow x - 1 = 56 - 33`

`\Rightarrow x - 1 = 23`

`\Rightarrow x = 23 + 1`

`\Rightarrow x = 24`

Vậy, `x = 24`

`c)`

`5 \times (4 + 6 \times x) = 290`

`\Rightarrow 4 + 6x = 290 \div 5`

`\Rightarrow 4 + 6x = 58`

`\Rightarrow 6x = 58 - 4`

`\Rightarrow 6x = 54`

`\Rightarrow x = 54 \div 6`

`\Rightarrow x = 9`

Vậy, `x = 9`

`d)`

`x. 3,7 + x. 6,3 = 120`

`\Rightarrow x . (3,7 + 6,3) = 120`

`\Rightarrow x . 10 = 120`

`\Rightarrow x = 120 \div 10`

`\Rightarrow x = 12`

Vậy,` x = 12.`

$#KDN040510$

\(BC\left(9,12\right)=\left\{0;36;72;...\right\}\)

\(BC\left(6,10\right)=\left\{0;30;60;...\right\}\)

\(BC\left(8,12\right)=\left\{0;24;48;...\right\}\)

\(BC\left(15,20\right)=\left\{0;60;120;...\right\}\)

\(BCNN\left(8,1\right)=8\)

\(BCNN\left(36,72\right)=36\cdot2=72\)

\(BCNN\left(60;150\right)=300\)

\(BCNN\left(10,12,15\right)=60\)

B( 9 ) = { 0 ; 9 ; 18 ; 27 ; 36 ; 45 ; 54; 63 ; 72; .... }

B( 12 ) = { 0 ; 12 ; 24 ; 36 ; 48 ; 60 ; 72 ; ...}

⇒ BC( 9, 12 ) = { 0 ; 36 ; 72 ;...}

B( 6 ) = { 0 ; 6 ; 12 ; 18 ; 24 ; 30 ; 36 ; 42 ; 48 ; ..}

B ( 10 ) = { 0 ;  10 ; 20 ; 30 ; 40 ..}

⇒ BC (6,10 ) =  { 0 ; 30 ... }

B ( 8 ) = { 0 ; 8 ; 16 ; 24 ; 32 ; 40 ; 48 ; 56, ...}

( Bội của 12 đã liệt kê bên trên )

⇒ BC ( 8,12 ) = { 0 ; 24 ;  36 ; 48 ; ...}

\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)

\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)

Ta có

\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)

\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)

\(\Rightarrow C>D\)

Ta có:

\(40=5\cdot2^3\)

\(50=5^2\cdot2\)

\(90=2\cdot3^2\cdot5\)

\(\RightarrowƯCLN\left(40,50,90\right)=5\cdot2=10\)

_________________

Ta có:
\(175=5^2\cdot7\)

\(250=2\cdot5^3\)

\(\RightarrowƯCLN\left(175,250\right)=5^2=25\)

_______________

Ta có:

\(100=2^2\cdot5^2\)

\(120=2^3\cdot5\cdot3\)

\(\RightarrowƯCLN\left(100,120\right)=2^2\cdot5=20\)

40 = 2³ . 5

50 = 2 . 5²

90 = 2 . 3² . 5

⇒ ƯCLN ( 40 , 50, 90 ) = 2 . 5 = 10.

\(\dfrac{2x+3}{x}=2+\dfrac{3}{x}\)

Để \(2x+3⋮x\Rightarrow3⋮x\Rightarrow x=\left\{-3;-1;1;3\right\}\)

56 = 2³ . 7

72 = 2³ . 3²

120 = 2³ . 3 . 5

⇒ ƯCLN ( 56 , 72 , 120 ) = 2³ = 8

CHO MIK XIN CÂU TRL NHANH NHÉ:)))))