a) Trong 1h, xe X đi nhanh hơn xe Y: \(50-40=10\left(km\right)\)

=> Thời gian cần để xe X cách 2km xe Y = 1/10 (h) = 6 phút

b) 3 phút trước 2 xe cách nhau 500m. Bây giờ 2 xe cách nhau 1km

=> Cứ 3 phút khoảng cách tăng thêm là 500m

=> Thời gian cần để xe X cách 2km xe Y = 3*2 = 6 phút

`63.(-25)+25.(-37)`

`=(-63).25+25.(-37)`

`=25.[(-63)+(-37)]`

`=25.(-100)`

`=-2500`

\(63\times\left(-25\right)+25\times\left(-37\right)\\ =63\times\left(-1\right)\times25+25\times\left(-37\right)\\ =25\times\left(-63-37\right)\\ =25\times\left(-100\right)\\ =-2500\)

`A=1+2+2^2+2^3+2^4+...+2^{200}`

`=>2A=2+2^2+2^3+2^4+2^5+...+2^{201}`

`=>2A-A=(2+2^2+2^3+2^4+2^5+...+2^{201})-(1+2+2^2+2^3+2^4+...+2^{200})`

`=>A=2^{201}-1`

`=>A+1=2^{201}`

\(A=5^2+5^4+5^6+...+5^{100}+5^{102}\\ =5^2.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)⋮25\)

\(=\left(5^2.1+5^2.5^2+5^2.5^4+....+5^2.5^{98}+5^2.5^{102}\right)\\ =5^2.\left(1+5^2+5^4+....+5^{98}+5^{102}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{102}\right)⋮25\\ =>A⋮25\)

\(\dfrac{1}{2}.x+\dfrac{3}{5}.x=-\dfrac{2}{3}\\ =>x.\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\\ =>x.\left(\dfrac{5+3.2}{10}\right)=-\dfrac{2}{3}\\ =>\dfrac{11}{10}x=-\dfrac{2}{3}\\ =>x=\left(-\dfrac{2}{3}\right):\dfrac{11}{10}\\ =>x=-\dfrac{20}{33}\)

đề là gì vậy ạ

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

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

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

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

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

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.

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

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

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

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

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

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

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