He's just gone out for a few minutes. (POP)

-> He's just popped out for a few minutes.

Chọn D

Chọn B

a. \(C=\dfrac{n+1}{n-2}\) \(\left(n\ne2\right)\)

\(C=\dfrac{n-2+3}{n-2}=\dfrac{n-2}{n-2}+\dfrac{3}{n-2}=1+\dfrac{3}{n-2}\)

Để C nguyên thì \(\dfrac{3}{n-2}\in Z\) \(\Leftrightarrow n-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

`@n-2=1->n=3(n)`

`@n-2=-1->n=1(n)`

`@n-2=3->n=5(n)`

`@n-2=-3->n=-1(n)`

Vậy \(n\in\left\{3;1;5;-1\right\}\) thì C nguyên

b.\(D=\dfrac{2n+1}{5n-3}\left(n\ne\dfrac{3}{5}\right)\)

Ta có: \(2n+1⋮5n-3\)

\(\Leftrightarrow5.\left(2n+1\right)⋮\left(5n-3\right)\)

\(\Leftrightarrow10n+5⋮5n-3\)

\(\Leftrightarrow2\left(5n-3\right)+11⋮\left(5n-3\right)\)

Vì \(2\left(5n-3\right)⋮\left(5n-3\right)\) nên để D nguyên thì  \(11⋮\left(5n-3\right)\) 

hay \(5n-3\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

`@5n-3=1->n=14/5(l)`

`@5n-3=-1->n=2/5(l)`

`@5n-3=11->n=14/5(l)`

`@5n-3=-11->n=-8/5(l)`

Vậy không có giá trị \(n\in Z\) thỏa mãn

Lọ 1 : AgNO₃

_{\(AgNO_3 + KI \to AgI + KNO_3\\ AgNO_3 + HCl \to AgCl + HNO_3\\ 2AgNO_3 + Na_2CO_3 \to 2NaNO_3 + Ag_2CO_3\)}

Lọ 2 : KI

Lọ 3 : Na₂CO₃

\(Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O\\ \)

Lọ 4 : HCl

He is always leaving his pen at home.

3/5:x=1/3

x=3/5:1/3

x=9/5

\(\frac{3}{5}\div x=\frac{1}{3}\)

\(x=\frac{3}{5}\div\frac{1}{3}=\frac{9}{5}\)

3/5:x=1/3

       x=3/5:1/3

       x=3/5×3/1

         x=9/5

vậy x=9/5

Không có bảng à sao mà làm

\(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\\ n_{AgCl}=n_{NaCl}=0,5.2=1\left(mol\right)\\ \Rightarrow m_{AgCl}=1.143,5=143,5\left(g\right)\\ \Rightarrow ChọnA\)