\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{36}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{4}{y}=\dfrac{5}{9}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{18}\\\dfrac{1}{x}=\dfrac{1}{36}-\dfrac{1}{18}=-\dfrac{1}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=18\\x=-36\end{matrix}\right.\)

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

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

He is always leaving his pen at home.

87 - 63 = 24

87 - 63 = 24 HT

87-63=24 nha em

Xl vì đã đăng linh tinh ạ

Cô e cho đề hay sao mà số lớn v

Hông bt nx

vui như tết

vui như tết

okok cảm ơn