$11$ giờ $30$ phút $=$ $11,5$ giờ

Gọi quãng đường $AB$ cần tìm là $x\left(km\right);(x>0)$

Thời gian dự định để đến $B$ là $11,5-7=4,5$ giờ

Thời gian đến $B$ theo thực tế là $12-7=5$ giờ

Vận tốc dự tính đến $B$ là $\frac{x}{4,5}\left(h\right)$

Vận tốc đến $B$ theo thực tế là $\frac{x}{5}\left(h\right)$

Theo đề , ta có phương trình:

$\frac{x}{4,5}-\frac{x}{5}=5$

C - B - A - D - D

skating

detests

swimming

listening

collection

had started
had lived
had eaten
had listened
had left
had finished
had been
had met
had forgot
hadn't rained
had not paid
had not done
had not eaten
had not brought
had not visited
had not met
had not had
had not used

B

D

B

A

C

A

D

C

$=\left(x^2+2x-1-2x\right)\left(x^2+2x-1-x\right)$

$=\left(x^2-1\right)\left(x^2+x-1\right)$

$=\left(x-1\right)\left(x+1\right)\left(x^2+x-1\right)$

$\iff {\left(2x+3\right)}^2-{\left(x-2\right)}^2=0$

$\iff \left(2x+3+x-2\right)\left(2x+3-x+2\right)=0$

=>(3x+1)(x+5)=0

=>x=-5 hoặc x=-1/3

went
had finished
had already seen
went
had not eaten
had already begun
watched 
had done
had he done
had gone
sat
watched
had gone
went
won

was reading

were you doing

were playing

were studying

was watching

came

\(=16x^m.\dfrac{1}{4}x^n+16x^m.\dfrac{3}{8}x^2+16^m.x^{m+2}\)

\(=4^{m+n}+6x^{m+2}+16x^{2m+2}\)

Vì tổng số hạt trong nguyên tử của một nguyên tố $Z$ là $40$

     $\Rightarrow 2p+n=40$ $(p=e)$   $\left(1\right)$

Vì số hạt mang điện nhiều hơn số hạt ko mang điện là $12$

     $\Rightarrow 2p-n=12$ $(p=e)$    $\left(2\right)$

Từ $\left(1\right);\left(2\right)\Rightarrow \{\begin{array}{c}p=13=e\\ n=14\end{array}$

Vậy nguyên tố $Z$ có cầu tạo là $p=e=13$ hạt; $n=14$ hạt