\(\left(1+2+3+...+2017\right)\times\left(1717\times18-1818\times17\right)\\ =\left(1+2+3+...+2017\right)\times\left(17\times101\times18-18\times101\times17\right)\\ =\left(1+2+3+...+2017\right)\times0\\ =0\)

`(-1/27) . 3/7 + 5/9 . (-3/7)`

`1/27 . (-3/7) + 5/9 . (-3/7)`

`(1/27 + 5/9) . (-3/7)`

`16/27 . (-3/7)`

`-16/63`

(\(\dfrac{3}{7}\)+(\(-\dfrac{3}{7}\))). \(\left(-\dfrac{1}{27}\right)\).\(\dfrac{5}{9}\)

= 0.\(\left(-\dfrac{1}{27}\right)\).​\(\dfrac{5}{9}\)

=0

\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}\)

=\(\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)\)

= \(0,9+\left(\dfrac{2}{6}+\dfrac{1}{6}\right)+\left(\dfrac{25}{35}-\dfrac{4}{35}\right)\)

= \(0,9+\dfrac{3}{6}+\dfrac{21}{35}\)

= `0,9 +0,5 + 0,6`

= `2`

\(\dfrac{2}{3}-\left[-\dfrac{7}{4}-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}+\dfrac{7}{4}+\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\)

\(=\dfrac{8}{12}+\dfrac{21}{12}+\dfrac{6}{12}+\dfrac{3}{8}\)

\(=\dfrac{35}{12}+\dfrac{3}{8}=\dfrac{70}{24}+\dfrac{9}{24}=\dfrac{79}{24}\)

\(\dfrac{2}{3}-\left[\dfrac{-7}{4}-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\\ =\dfrac{2}{3}-\left[\dfrac{-7}{4}-\left(\dfrac{4}{8}+\dfrac{3}{8}\right)\right]\\ =\dfrac{2}{3}-\left(\dfrac{-7}{4}-\dfrac{7}{8}\right)\\ =\dfrac{2}{3}-\left(\dfrac{-14}{8}-\dfrac{7}{8}\right)\\ =\dfrac{2}{3}+\dfrac{21}{8}\\ =\dfrac{16}{24}+\dfrac{63}{24}\\ =\dfrac{79}{24}\)

k+10<25

=>k<15

x\(\in\)N^*

=>x>0

=>4k+2>0

=>k>-0,5

=>-0,5<k<15

mà k nguyên

nên k\(\in\left\{0;1;2;3;...;14\right\}\)

=>\(x\in\left\{2;6;10;...;58\right\}\)

Đê bài sai rồi nhé bạn!
Để một số chai hêt scho 5 thì số đó phải kết thúc bằng 0 hoặc 5. Mà số đó có 2 chữ số giống nhau nên số đó là 55, nhưng 55 chia cho 2 dư 5 (k thỏa mãn đề bài)

480:6x3+25,3-24448:24

=80x3+25,3-3056/3

\(=240+25,3-\dfrac{3056}{3}=-\dfrac{22601}{30}\)

\(480:6\times3+25,3-24448:24\)

\(=80\times3+25,3-\dfrac{3056}{3}\)

\(=240+25,3-\dfrac{3056}{3}\)

\(=\dfrac{2653}{10}-\dfrac{3056}{3}\)

\(=-\dfrac{22601}{30}\)

Bài đúng nhưng mình chỉnh sửa cho bạn phần trình bày nhé (bạn có thể thay đổi hoặc không).

Theo đề bài ta có:
\(\left(a+b\right)\) chia hết cho \(m\) và \(a\) chia hết cho \(m\) 
\(\Rightarrow\left(a+b\right)=x.m;a=y.m\left(x,y\inℕ;x\ge y\right)\)
\(\Rightarrow\left(a+b\right)-a=x.m-y.m\\ \Rightarrow b=m.\left(x-y\right)\)
Vì \(m.\left(x-y\right)⋮m\) nên \(b⋮m\)
Vậy...


 

\(450=2\cdot3^2\cdot5^2;354=2\cdot3\cdot59\)

=>\(ưcln\left(450;354\right)=3\cdot2=6\)

=>\(ƯC\left(450;354\right)=Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Ta có :

\(450\times2\times3^2\)\(\times5^2\)

\(354=2\times3\times59\)

\(\RightarrowƯCLN\left(450,354\right)\)\(=2\times3\)\(=6\)

\(\RightarrowƯC\left(450,354\right)\)\(=Ư\left(6\right)\)\(\left\{-1;1;2;-2;3;-3;6;-6\right\}\)