19 - 42 - 19 - 57 - 19

= -23 - 19 - 57 - 19

= -42 - 57 - 19

= -99 - 19

= -118

19 - 42(-19) - 57(-19) = 19 - ( 42 + 57 )(-19)

                                   = 19 - 99(-19)

                                   = 19 - (-1881) = 19 + 1881 = 1900

a) 33890 = 10 x 3389 = 2 x 5 x 3389

38002 = 2 x 19001

b) 31798 = 26 x 1223 = 2 x 13 x 1223

6621 = 3 x 2207

a) 33890 = 10 x 3389 = 2 x 5 x 3389

38002 = 2 x 19001

b) 31798 = 26 x 1223 = 2 x 13 x 1223

6621 = 3 x 2207

\(\left(42-69+17\right)-42+17\\ =42-69+17-42-17\\ =\left(42-42\right)+\left(17-17\right)-69\\ =0+0-69\\ =-69.\)

( 42 - 69 + 17 ) - ( 42 + 17 )

= 42 - 69 + 17 - 42 - 17

= ( 42 + 17 - 42 - 17 ) + 69

= 0 + 69 

= 69

Xem lại đề.

5 . 3^x+2 - 10 = 125

5 . 3^x+2 = 125 + 10

5. 3^x+2 = 135

3^x+2 = 135 : 5 

3^x+2 = 27

3^x+2 = 3^3

x+2 = 3

x = 3 - 2

x = 1

Vậy x = 1

a,  24 ⋮ (2n + 1)

     2n + 1 \(\in\) Ư(24) = {1; 2; 3; 4; 6; 8; 12; 24}

     n \(\in\) {0; \(\dfrac{1}{2}\); 1; \(\dfrac{3}{2}\); \(\dfrac{5}{2}\); \(\dfrac{7}{2}\); \(\dfrac{11}{2}\); \(\dfrac{23}{2}\)}

    Vì n \(\in\) N nên n \(\in\) {0; 1}

b, n - 3 \(\in\) Ư(8) 

     n - 3 \(\in\) {-8; -4; -2; -1; 1; 2; 4; 8}

     n       \(\in\) {-5; -1; 1; 2; 4; 5; 7; 11}

Vì n \(\in\) N nên n \(\in\) {1; 2; 4;5;7;11}

\(19-\left[179-\left(24.10-2^4.5\right)\right]\\ =19-\left[179-\left(240-16.5\right)\right]\\ =19-\left[179-\left(240-80\right)\right]\\ =19-\left[179-160\right]\\ =19-19\\ =0.\)

likes r ạ

(-199) + (-200) + (-201)

=-(199+200+201)

=-600

camon ạ

A=-418-{-218-[-118-(-318)+2012]}

A=-418+218+118+318+2012

A=2248

A= - 418 - {-218-[-118-(-318)+2012]}

A= - 418 - {-218-[-118-(-318)+2012]}

A= - 418 - {-218-[-118+318+2012]}

A= - 418 - {-218+118-318-2012}

A= - 418 + 2018 - 118 + 318 + 2012

A= 1600 - 118 + 218 + 2012

A= 1582 + 218 + 2012

A= 1800 + 2012

A= 3812

mình đang cần gấp vì mai khảo sát hs giỏi nên mọi người giúp

1.

$(a+b+c)-(a-b+c)=a+b+c-a+b-c=(a-a)+(b+b)+(c-c)=0+2b+0=2b$

2.

$(a-b+c)-(a-b+c)=0$

3.

$(a+b+c)-(b-a+c)=a+b+c-b+a-c=(a+a)+(b-b)+(c-c)=2a+0+0=2a$

4.

$(a-c)-(d+b+a+c)=a-c-d-b-a-c=(a-a)+(-c-c)-d-b=0-2c-d-b=-2c-d-b$

5.

$(a+d-c)-(a+b-c)=a+d-c-a-b+c=(a-a)+(-c+c)+d-b=0+0+d-b=d-b$

6.

$(a-b+c+d)+(a+c-d-b)=a-b+c+d+a+c-d-b$

$=(a+a)+(-b-b)+(c+c)+(d-d)=2a-2b+2c$

7.

$(a+d-c)+(a-b+c)=a+d-c+a-b+c=(a+a)+d+(-c+c)=2a+d$