Bài 13: 

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

hay x=0

cần kết quả đúng ko 

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

`a)|2x-15|=13`

`**2x-15=13`

`<=>2x=28`

`<=>x=14.`

`**2x-15=-13`

`<=>2x=-2`

`<=>x=-1.`

`b)|7x+3|=66`

`**7x+3=66`

`<=>7x=63`

`<=>x9`

`**7x+3=-66`

`<=>7x=-69`

`<=>x=-69/7`

`c)|5x-2|=0`

`<=>5x-2=0`

`<=>5x=2`

`<=>x=2/5`

\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

Vậy ...

\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

Vậy ...

\(c,\Leftrightarrow5x-2=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

a \(\Rightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x=18\\2x=-8\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

b \(\Rightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}7x=63\\7x=-69\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)

c \(\Rightarrow5x-2=0\Rightarrow x=\dfrac{2}{5}\)

a  21-4x=13
  -4x=13-21
-4x=-8
x=-8/-4
x=2
 

b x=5

  chúc bạn học tốt

c: Ta có: 11+x:5=13

\(\Leftrightarrow x:5=2\)

hay x=10

d: Ta có: \(13+2\left(x+1\right)=15\)

\(\Leftrightarrow2x+2=2\)

\(\Leftrightarrow2x=0\)

hay x=0

e: Ta có: 2x+21=41

\(\Leftrightarrow2x=20\)

hay x=10

f: Ta có: \(12+3\left(x-2\right)=60\)

\(\Leftrightarrow3\left(x-2\right)=48\)

\(\Leftrightarrow x-2=16\)

hay x=18

g: Ta có: \(24x-11\cdot13=11\cdot11\)

\(\Leftrightarrow24x=11\cdot24\)

hay x=11

h: Ta có: \(17-\left(x-4\right):2=3\)

\(\Leftrightarrow\left(x-4\right):2=14\)

\(\Leftrightarrow x-4=28\)

hay x=32

Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{-3}{4}\)

⇒\(\dfrac{x}{-3}=\dfrac{y}{4}\) 

⇒\(\dfrac{2x}{-6}=\dfrac{3y}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{-6}=\dfrac{3y}{12}=\dfrac{3y-2x}{12-\left(-6\right)}=\dfrac{36}{18}=2\)

⇒\(\left\{{}\begin{matrix}x=2.-3=-6\\y=2.4=8\end{matrix}\right.\)

a) 70 - 5(x - 3 ) = 45

5( x - 3 ) = 70 - 45 = 25

x - 3 = 25 : 5 = 5

x = 5 + 3 = 8

b) (2x - 1 )⁴ = 3 . 6² - 27

(2x - 1 )⁴ = 3 . 36 - 27

(2x - 1 )⁴ = 81

 Ta thấy 81  = 3⁴ vậy suy ra (2x - 1)⁴ = 3⁴

 Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2

Thử lại : 2 . 2 - 1 = 4 - 1 = 3

Vậy x = 2

c) 3x³ +  43 = 10² - 33

3x³ + 43 = 100 - 33 = 67

3x³ = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)