nH2 = 4.704/22.4 = 0.21 (mol) 

2X + 2nHCl => 2XCln + nH2 

0.42/n_______________0.21 

M_X = 3.78/0.42/n = 9n 

BL : n = 3 => X là : Al 

nH₂ = 0,21 (mol/0

pt: 2X + 2nHCl \(\rightarrow\) 2XCl_n + nH₂

     \(\dfrac{3,78}{X}\)                             0,21

Theo pt: \(\dfrac{3,78}{X}=\dfrac{0,42}{n}\)

=> 3,78n = 0,42X

=> \(\dfrac{X}{n}=9\)

Do X là kim loại => X có hoá trị n = I, II, III

Thử từng giá trị của n => n = 3 => X là Al

`1)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)`

`<=>2x^2-5x-12+x^2-7x+10=3x^2-17x+20`

`<=>3x^2-12x-2=3x^2-17x+20`

`<=>5x=22`

`<=>x=22/5`

Vậy `S={22/5}`

`2)x^2(x-2019)=2019-x`

`<=>(x-2019)(x^2+1)=0`

`<=>x-2019=0`

`<=>x=2019(do \ x^2+1>=1>0)`

Vậy `S={2019}`

x=14

\(2 + 4 + 6 + 8 + ... + 2x = 210\\ [\frac{2x-2}{2} + 1 ] . \frac{2x+2}{2} = 210\\ [ \frac{2x-2.1}{2} + 1 ] .\frac{2x+2.1}{2} = 210\\ [\frac{2(x-1)}{2} + 1 ] .\frac{2(x+1)}{2} = 210\\ [ ( x -1 ) + 1 ] . ( x + 1 ) = 210\\ x . ( x + 1 ) = 210\\ x . ( x + 1 ) = 14 . 15\\ \Rightarrow x = 14\\\)

x=14

Cho xin một like đi các dân chơi à.

20x - 5x = 0

15x = 0

x = 0

X=6

cần gấp lắm ạ xin các bạn giúp đỡ . ___ .

Câu 1: 

0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3

= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059

= 9 x 43,6 + 9 x 85,8 + 6 x 105,9

= 3 x 130,8 + 3 x 257,4 + 3 x 211,8

= 3 x ( 130,8 + 257,4 + 211,8 )

= 3 x 600 

= 1800

Câu 2: 

3/4 x X + 1/2 x X - 15 = 35

X x ( 3/4 + 1/2 ) - 15 = 35

X x ( 3/4 + 1/2 ) = 50

X x 5/4 = 50

X = 40

VẬy X = 40

1.  0.9x218x2+0.18x4290+0.6x353x3  

= 1.8x218+0.18x10x429+1.8x353

=1.8x218+1.8x429+1.8x353

= 1.8x( 218+429+353 )

= 1.8x1000

=1800

2. 

3/4 x X + 1/2 x X - 15 = 35

3/4 x X + 1/2 x X = 35+15

3/4 x X +1/2 x X = 50

X x ( 3/4 + 1/2 ) = 50 

X x 5/4 = 50 

X = 50 : 5/4

X =40

1.

=3/5x(3/7+4/7)+2/5x(13/9-4/9)

=3/5x1+2/5x1

=3/5+2/5

=1

2.Xx(3/4+4/5)=7/10

Xx31/20=7/10

X         =7/10:31/20

X         =14/31

\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)

\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)

\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)

_________________________________________

\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)

\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)

\(\frac{31}{20}\cdot x=\frac{7}{10}\)

\(x=\frac{7}{10}:\frac{31}{20}\)

\(x=\frac{14}{31}\)

\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x^2}{9-x^2}\)     ĐKXĐ : \(x\ne\pm3\)

\(\Leftrightarrow\frac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{3x^2-7x^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2=3x^2-7x^2\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(-3x^2-x^2-3x^2-3x^2+7x^2\right)-3x=0\)

\(\Leftrightarrow-3x^2-3x=0\)

\(\Leftrightarrow-3x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

KL : nghiệm của PT là : \(S=\left\{0;-1\right\}\)

\(\frac{x-4}{x-1}+\frac{x+4}{x+1}=2\) DKXĐ : \(x\ne\pm1\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=2\)

\(\Leftrightarrow x^2+x-4x-4+x^2-x+4x-4=2\)

\(\Leftrightarrow\left(x^2+x^2\right)\left(x-4x-x+4x\right)+\left(-4-4\right)=2\)

\(\Leftrightarrow2x^2-8=2\)

\(\Leftrightarrow2x^2=10\)

.....

x = 0+1+2+3+4

x=10

x = 0  + 4 + 3 + 2 + 1

x = 10