$\dfrac{3}{7}$ $=$ $\dfrac{2x + 1}{3x + 5}$

$\Rightarrow$ $\text{7.(3x + 5) = 3.(2x + 1)}$

$\Rightarrow$ $\text{21x + 35 = 6x + 3}$

$\Rightarrow$ $\text{15x = -32}$

$\Lo ...

1.A

 Vì $\dfrac{6}{30}$ = $\dfrac{1}{5}$ = 0,2
2.C
Vì $\dfrac{14^3}{7^3}$ = $\dfrac{2744}{343}$ = 8

3.D

Ta có : $n_{BaO}$ = $\dfrac{53,55}{153}$ = 0,35 (mol)
           $n_{H2SO4}$ = $\dfrac{300 . 14,6%}{100%}$ ≈ 0,4

        & ...

$CO_2$ + NaOH → $Na_2$$CO_3$ + $H_2$O
a)Nếu V = 2,24
→ n $CO_2$ = $\frac{2,24}{22,4}$ = 0,1 (mol)

Ta có : n NaOH = $\frac{200}{1000}$ . 2 = 0,4 (mol)
Thấy n N ...

$\text{$M_2$$O_y$ + y$H_2$$SO_4$ → $M_2$$(SO4)_y $ + y$H_2$O}$
$\text{Đặt n M2Oy = 1 (mol)}$
$\text{⇒ m M2Oy = 2M + 16y (g) }$
$\text{Theo pt : nH2SO4 = y n M2Oy = y (mol) ...

$\dfrac{13}{3}$ `-` $\dfrac{21}{4}$ `+` $\dfrac{23}{6}$ `-` $\dfrac{31}{12}$
`=` $\dfrac{52}{12}$ `-` $\dfrac{63}{12}$ `+` $\dfrac{46}{12}$ `-` $\dfrac{31}{12}$
`=` $\dfrac{52 - 63 + ...

$\text{Áp dụng định lý Pytago vào ΔABC ($\widehat{H}$ = $90^\circ$) có :}$

$\text{$AB^2$ = $AH^2$ + $HB^2$}$

$\text{⇔ $HB^2$ = $5^2$ - $4^2$}$

TBR,Ta có :

`x` ∈ Ư(90) = {1;2;3;5;6;9;10;15;18;30;45;90}

`x` ∈ Ư(15) = {1;3;5;15}

Mà ƯC(90,15) = {1 ; 2 ; 3 ; 5}

Mà 3 < `x` < 10

Gọi số học sinh cần tìm là $x$ ($x$ ∈ $N*$)

$\text{Vì khi xếp môi xe 35,42,45 học sinh đều vừa đủ}$

$\text{⇒ $x$ ∈ BC(35,42,45)}$

$\te ...

$\text{100 + 22 + 33 + 11 + 100 + 33 + 22}$

$\text{= (100 + 100) + (22 + 33) + (33 + 22) + 11}$

$\text{= 200 + 55 + 55 + 11}$

$\text{= 200 + 110 + 11}$

$\text{= 310 + 11 ...