`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

⇔\(7\left(x-3\right)=5\left(x+5\right)\)

⇔\(7x-21=5x+25\)

⇔\(7x-21-5x-25=0\)

⇔\(2x-46=0\)

⇔\(2x=46\)

⇔\(x=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

⇔\(\left(x+1\right)\left(x-1\right)=7.9\)

⇔\(x^2-1=63\)

⇔\(x^2=64=8^2\)

⇔\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(a.\dfrac{x-3}{x+5}=\dfrac{5}{7}\)                          

    \(\left(x-3\right).7=\left(x+5\right).5\) 

    \(\left(x.7\right)+\left[\left(-3\right).7\right]=\left(x.5\right)+\left(5.5\right)\)

    \(7x-21=5x+25\)

    \(7x-5x=25+21\)

             \(2x=46\)

               \(x=46:2\)

               \(x=23\)

Câu b/ cứ làm theo câu a/ là được

AcCl₃ và NiCl₂

Bạn ơi bạn làm đc bao nhiêu thì làm phụ mình nhé 

bài1

a, 

5 - 7 + 3 +(-8)

= -2 + 3 + ( - 8)

=     1    + (-8)= -7

a; - (-5) - (+7) + (+3) + (-8)

  = 5 - 7 + 3  - 8

= (5 + 3 - 8) - 7

= (8 - 8) - 7

= 0 - 7

= - 7

       4/3 x 5/4 x 6/5 x 7/6 x  8/7 x 9/8 

   =  4/9   

1. -x+20 = -(-15)-8+13

=> -x=15-8+13-20

=> -x=0

=> x=0

2. -(-10)+x=-13+(-9)+(-6)

=> 10+x=-13-9-6

=> x = -13-9-6-10

=> x = -38

3. 8-(-12)+10=-(-14)-x

=> 8+12+10=14-x

=> x = 14-8-12-10

=> x = -16

4. -(+12)+(-x)-(-3)=5-(-7)

=> -12-x+3=5+7

=> -x=5+7+12-3

=> -x=21

=> x=-21

5. 14-x+(-10)=-(-9)+(+15)

=> 14-x-10=9+15

=> -x=9+15-14+10

=> -x=20

=> x=-20

6. 12-(-17)+(-3)=-5+x

=> 12+17-3+5=x

=> x=31

7. x-(-19)-(+32)=14-(+16)

=> x+19-32=14-16

=> x=14-16+32-19

=> x=11

8. x-|-15|-|7|=-(-9)+|-5|

=> x-15-7=9+5

=> x=9+5+7+15

=> x=36

9. 15-x+17=13-(-21)

=> 15-x+17=13+21

=> -x=13+21-15-17

=> -x=2

=> x=-2

10. -|-5|-(-x)+4=3-(-25)

=> -5+x+4=3+25

=> x=3+25-4+5

=> x=29

a)

\(\frac{x-3}{10}=\frac{4}{x-3}\)

=> ( x - 3 )² = 4 . 10.

     ( x - 3 )² = 40

Mà x - 3 thuộc Z ( vì x thuộc Z ) nên ( x - 3 )² là số chính phương.

Do 40 không là số chính phương.

=> Ko tìm được x thuộc Z thỏa mãn đề bài.

b) 

\(\frac{x+5}{9}=\frac{4}{x+5}\)

=> ( x + 5 )² = 4 . 9

     ( x + 5 )² = 36

=> x + 5 = 6 hoặc x + 5 = -6.

+) x + 5 = 6

           x = 1.

+) x + 5 = -6

          x = -11.

Vậy x = 1; x = -11.

c)

\(\frac{x}{4}=\frac{18}{x+1}\)

=> x ( x + 1 ) = 4 . 18

     x ( x + 1 ) = 72

Mà x và x + 1 là các số nguyên. ( do x thuộc Z )

=> x và x + 1 là các ước nguyên của 72.

Các ước nguyên của 72 là: -1 ; -2 ; -3 ; -4 ; -6 ; -8 ; -9 ; -12 ; -18 ; -36 ; -72 ; 1 ; 2 ; 3 ; 4 ; 6 ; 8 ; 9 ; 12 ; 18 ; 36 ; 72 .

Mặt khác, x và x + 1 là 2 số nguyên liên tiếp và x + 1 > x.

=> x = 8 ; x + 1 = 9

hoặc x = -9 ; x + 1 = -8.

Vậy  x = 8 ; x = -9.

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)