\(-\dfrac{3}{5}\times x+\dfrac{1}{2}=\dfrac{4}{5}\)

\(-\dfrac{3}{5}\times x=\dfrac{4}{5}-\dfrac{1}{2}\)

\(-\dfrac{3}{5}\times x=\dfrac{3}{10}\)

\(x=\dfrac{3}{10}:-\dfrac{3}{5}\)

\(x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\)

\(\dfrac{1}{2}\times x+\dfrac{3}{8}=\dfrac{7}{16}\)

\(\dfrac{1}{2}\times x=\dfrac{7}{16}-\dfrac{3}{8}\)

\(\dfrac{1}{2}\times x=\dfrac{1}{16}\)

\(x=\dfrac{1}{8}\)

Vậy \(x=\dfrac{1}{8}\)

\(\dfrac{2}{3}x-\dfrac{5}{6}x=-\dfrac{1}{2}\)

\(x\times\left(\dfrac{2}{3}-\dfrac{5}{6}\right)=-\dfrac{1}{2}\)

\(x\times\dfrac{-1}{6}=-\dfrac{1}{2}\)

\(x=3\)

Vậy \(x=3\)

\(\dfrac{2}{5}x+\dfrac{3}{10}x=-\dfrac{1}{5}\)

\(x\times\left(\dfrac{2}{5}+\dfrac{3}{10}\right)=-\dfrac{1}{5}\)

\(x\times\dfrac{7}{10}=-\dfrac{1}{5}\)

\(x=-\dfrac{2}{7}\)

Vậy \(x=-\dfrac{2}{7}\).

Lời giải:
$\frac{53}{17}=\frac{51+2}{17}=\frac{51}{17}+\frac{2}{17}=3+\frac{2}{17}$

$>3+\frac{2}{23}=\frac{3.23+2}{23}=\frac{71}{23}$

Đề bị lỗi. Bạn coi lại đề.

\(n^2+13n-13=\left(n^2+3n\right)+\left(10n+30\right)-43\\ =n\left(n+3\right)+10\left(n+3\right)-43\\ =\left(n+3\right)\left(n+10\right)-43\)

\(Để:n^2+13n-13⋮\left(n+3\right)\\ =>43⋮\left(n+3\right)\\ =>n+3\inƯ\left(43\right)=\left\{\pm1;\pm43\right\}\\ =>n\in\left\{-4;-2;-46;40\right\}\left(TMDK\right)\)

\(2.B=\dfrac{2}{6}+\dfrac{2}{14}+\dfrac{2}{60}+...+\dfrac{2}{990}\)

\(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{9.10.11}\)

\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11}\)

\(2B=\dfrac{1}{1.2}-\dfrac{1}{10.11}\)

\(B=\dfrac{27}{110}\)

Với 1 số tự nhiên a bất kì \(\Rightarrow\left[{}\begin{matrix}a=3x\\a=3x+1\\a=3x+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a^2=9x^2\\a^2=9x^2+6x+1\\a^2=9x^2+12x+4\end{matrix}\right.\)

Tổng 2 số chính phương \(p=a^2,q=b^2\) chia hết cho 3 => \(p=9x^2,q=9y^2\Rightarrow p,q⋮9\)

\(\dfrac{31}{61}\) = 1 - \(\dfrac{30}{61}\)

\(\dfrac{41}{71}\) =  1  - \(\dfrac{30}{71}\)

vì \(\dfrac{30}{61}\) >  \(\dfrac{30}{71}\) nên \(\dfrac{31}{61}< \dfrac{41}{71}\)

31/61 = 2201/4331 

41/71 = 1271/4331

Vậy 31/61 > 41/71

phương pháp phản chứng:

giả sử ( 459a + 1035b) ⋮ 45 ∀ a, b \(\in\) N

Với a = 1; b = 1 ta có:

(4591 + 10351) ⋮ 45 ⇔ \(\overline{...1}\) ⋮ 45 ⇒ \(\overline{...1}\) ⋮ 5 (vô lý)

Vậy ( 459a + 1035b) ⋮ 45 ∀ a, b \(\in\) N  là không thể 

Bạn có chép sai đề bài không ạ?

= (-1+3) + ( -5 + 7)+...+ (97 - 99) 

= 2 + 2 + ...+ 2 

Ta có tất cả số cặp là ( (99-1):2+1):2=25 (cặp)

=> Tổng trên bằng: 25 x 2 = 50

Lời giải:

$-1+3-5+7-...-97+99$

$=(-1-5-9-...-97)+(3+7+...+99)$

$=(3+7+...+99)-(1+5+9+...+97)$

$=[(99-3):4+1].\frac{99+3}{2}-[(97-1):4+1].\frac{97+1}{2}$

$=1275-1225=50$