cái này thì triệt tiêu ra đó bn

là bằng 2020/6069

A = 1/3.5.7 + 1/5.7.9+ ... + 1/2019.2021.2023

2A = 2/3.5.7 + 2/5.7.9+ ... + 2/2019.2021.2023

2A = 1/3-1/5+1/7+1/5-1/7+1/9+....+1/2019-1/2021+1/2023

2A = 1/3 - 1/2023

2A = 2023/6069 - 3/6069

2A = 2023-3/6069

2A = 2020/6069

 A = 1010/6069

Vậy A = 1010/6069

\(5,\Leftrightarrow8\left(x^2-5x+3x-15\right)-\left(8x^2-4x+20x-10\right)=0\)

\(\Leftrightarrow8x^2-40x+24x-120-8x^2+4x-20x+10=0\)

\(\Leftrightarrow-32x-110=0\)

\(\Leftrightarrow x=\dfrac{110}{-32}=-\dfrac{55}{16}\)

\(9,\Leftrightarrow\left(5x-5\right)\left(2x+7-2x-1\right)=x+2-x+5\)

\(\Leftrightarrow6\left(5x-5\right)=7\)

\(\Leftrightarrow5x-5=\dfrac{7}{6}\)

\(\Leftrightarrow x=\dfrac{37}{30}\)

\(8,\Leftrightarrow\left(6x-3\right)\left(x-5-3x+1\right)-3+13=0\)

\(\Leftrightarrow-6\left(2x-1\right)\left(x+2\right)+10=0\)

\(\Leftrightarrow2x^2-x+4x-2=\dfrac{10}{6}\)

\(\Leftrightarrow2x^2+3x-\dfrac{11}{3}=0\)

\(10,\Leftrightarrow8x\left(x^2+x+6x+6\right)-8x^3=5x\)

\(\Leftrightarrow8x^3+8x^2+48x^2+48x-8x^3-5x=0\)

\(\Leftrightarrow56x^2+43x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{43}{56}\end{matrix}\right.\)

Vậy ....

;-; quên đọc đề may làm trúng 9; 10 câu 6 nốt đây nha

\(6,\Leftrightarrow9x^2-9x-8x^2+40-x^2=5x-3-7x+4\)

\(\Leftrightarrow9x^2-9x-8x^2+40-x^2-5x+3+7x-4=0\)

\(\Leftrightarrow-7x+39=0\)

\(\Leftrightarrow x=\dfrac{39}{7}\)

Vậy ...

1 He doesn't play games

2 I don't go to the art club

3 What about sitting down?

4 We are going to take part in the stamp collector's club

5 How about relaxing?

6 My favorite subject is English

7 My favorite sport is football

8 Pink and blue are my favorite colors

9 My favorite subject is English

10 My favorite sport is football

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\\ \Rightarrow\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}\)

Áp dụng t/c dtsbn:

\(\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}=\dfrac{a+b+c+d-a-b+c+d}{a-b+c-d-a+b+c-d}=\dfrac{2c+2d}{2c-2d}=\dfrac{c+d}{c-d}\)

\(\dfrac{a+b+c+d}{a-b+c-d}=\dfrac{a+b-c-d}{a-b-c+d}=\dfrac{a+b+c+d+a+b-c-d}{a-b+c-d+a-b-c+d}=\dfrac{2a+2b}{2a-2b}=\dfrac{a+b}{a-b}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

Áp dụng t/c dtsbn:

\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b-a+b}{c+d-c+d}=\dfrac{2b}{2d}=\dfrac{b}{d}\\ \dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\\ \Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\left(2+5+7\right)x=4\times7\)

\(14x=28\)

\(x=2\)

Trl :

2x + 5x + 7x = 2².7

14x = 4.7

14x = 28

x= 28 : 14

x = 2

Hok tốt

a ) H2SO4 + 2NaOH ----------> Na2SO4 + H2O

H2SO4 + 2NaHCO3 ----------> Na2SO4 + 2H2O + 2CO2

\(n_{H_2SO_4}=0,5a\)

\(n_{NaOH}=0,2.2=0,4\left(mol\right)\)

\(n_{NaHCO_3}=\dfrac{0,42}{84}=0,005\left(mol\right)\)

Trường hợp 1: H2SO4 dư
H2SO4 + 2NaOH ----------> Na2SO4 + H2O
0,2<---------0,4
n_H2SO4dư = 0,5a - 0,2 (mol)

=> \(\dfrac{1}{2}n_{H_2SO_4}=0,25a-0,1\left(mol\right)\)

H2SO4 + 2NaHCO3 ----------> Na2SO4 + 2H2O + 2CO2

\(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaHCO_3}=0,0025\left(mol\right)\)

=> \(0,25.a-0,1=0,0025\)

=> a=0,41 (M)

Trường hợp 2: NaOH dư
H2SO4 + 2NaOH -----> K2SO4 + 2H2O
0,5a-------->a
n_NaOHdư = 0.4 - a (mol)

=> \(\dfrac{1}{2}n_{NaOH\left(dư\right)}=0,2-0,5a\left(mol\right)\)
NaHCO3 + NaOH -------> Na2CO3 + H2O
0,005-------->0,005
=> 0,2 - 0.5a = 0,005

=> a = 0,39

Câu 1

1)

$(1) 4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
$(2) 2SO_2 + O_2 \xrightarrow{t^o,V_2O_5} 2SO_3$
$(3) SO_3 + H_2O \to H_2SO_4$

$(4)6FeSO_4 + 3Cl_2 \to 2Fe_2(SO_4)_3 + 2FeCl_3$

$(5) Fe_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2FeCl_3$

$(6) FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl$
$(7) 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$

2)

Trích mẫu thử

Cho dung dịch phenolphtalein vào

- mẫu thử hóa hồng là NaOH

Cho ba mẫu thử còn lai tác dụng lần lượt với nhau

- mẫu thử tạo kết tủa trắng là $H_2SO_4,Ba(NO_3)_2$
$Ba(NO_3)_2 + H_2SO_4 \to BaSO_4 + 2HNO_3$

- mẫu thử không hiện tượng là HCl

Cô cạn hai mẫu thử còn :

- mẫu thử thu được chất rắn khan là $Ba(NO_3)_2$

- mẫu thử không thu được chất rắn là $H_2SO_4$