$a)PTHH:$

  $S+O_2$ $\stackrel{t^o}{\to }$ $S{O}_2$

$0,05$ $0,05$               $0,05$        $\left(mol\right)$

$b)n_S=\frac{1,6}{32}=0,05\left(mol\right)$

$\circ V_{S{O}_2}=0,05.22,4=1,12\left(l\right)$

$(2xy-1)(3xy-2x+3y)$

$=2xy.3xy+2xy.(-2x)+2xy.3y-1.3xy-1.(-2x)-1.3y$

$=6x^2{y}^2-4x^{2}y+6x{y}^2-3xy+2x-3y$

Gọi thể tích dd HCl 36% cần dùng là a (lít)

$m_{dd.HCl.36\%}=1000.a.1,02=1020a\left(g\right)$

$\Rightarrow m_{HCl}=\frac{1020a.36}{100}=367,2a\left(g\right)\Rightarrow n_{HCl}=\frac{367,2a}{36,5}=\frac{3672}{365}a\left(mol\right)$

Mặt khác, n_HCl = 0,1.2 = 0,2 (mol)

=> a = $\frac{73}{3672}\left(lít\right)$

Thể tích nước cần dùng là: $0,1-\frac{73}{3672}=\frac{1471}{18360}\left(lít\right)$

a)a)Thế x=10x=10 vào AA, ta được:

A=1x−1=110−1=19A=1x−1=110−1=19

b)b)ĐK:x≠1ĐK:x≠1

S=B−AS=B−A

S=x2+2x3−1+x+1x2+x+1−1x−1

S=x2+2(x−1)(x2+x+1)+x+1x2+x+1−1x−1

S=x2+2+(x+1)(x−1)−(x2+x+1)(x−1)(x2+x+1)

S=x2+2+x2−1−x2−x−1(x−1)(x2+x+1)

S=x2−x(x−1)(x2+x+1)

S=xx2+x+1

c)c)

Ta có: (x−1)2≥0(đúng)

⇔x2−2x+1≥0

⇔x2+x+1≥3x

S=xx2+x+1≤x3x≤13( vì mẫu số càng lớn thì phân số càng nhỏ )

SKBr(10oC)=\(\dfrac{21,42}{36}.100=59,5\left(g\right)\)C%KBr(bão.hoà)=\(\dfrac{59,5}{100+59,5}\) \(.100\%\) = 37,3%

(1)<=> 3x=2x+1<=>x=1

(2) <=>x^2=7<=>x=$\sqrt{7};x=-\sqrt{7}$

(3)<=>x=7

(4)<=>x=8

(5)<=>x^2=9<=>x=3;x=-3

3x² - (3x - 3) (-2 + x) = 39

<=>3x^2-(-6x+3x^2+6-3x)=39

<=> 9x-6=39

<=>9x=45

<=>x=5

=(√a−2)(a+2√a+4)

C

D

B