`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

Bài 13: 

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

hay x=0

a)  (x - 3)² - 5.(x - 2) + 5 = 0.

<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0

<=> x^2 - 11x + 24 = 0

<=> (x-3)(x-8)=0

<=> x = 3 hoặc x = 8

b) (2x - 1)² - 3.(x - 2).(x + 2) - 25 = 0.

<=> 4x^2 - 4x + 1 - 3x^2 + 12 - 25 = 0

<=> x² - 4x - 12 = 0

<=> (x+2)(x-6) = 0

<=> x = -2 hoặc x = 6

d)  x² - 4x + 5 = 0.

<=> (x - 2)² = -1 (vô lý)

Vậy phương trình vô nghiệm

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

a,(2x-5^2)-4x(x-3)=0

=> 2x-25-4x²+12x=0

=>-4x²+14x-25=0

đề bài ý a sai nha

b, 6x²-7x=0

=>x(6x-7)=0

=>x=0 và 6x-7=0

=>x=0 và x=7/6

vậy x=0 và x=7/6

a, x + 15,96 = 17,485

x = 17,485 - 15,96 = 1,525

b, 99,99 - x = 5, 678

x = 99,99 - 5,678 = 94,312

c, x = 105,85 : 5 = 21,17

d, x = 9,8 x 8,9 = 87,22

e, x = 76,32 : 36 = 2,12

Bài 2: Tìm x, biết:

a) x + 15,96 = 1,345 x 13

x + 15,96 = 17,485 

x = 17,485 - 15,96

x = 1.525

b) 99,99 – x = 5,678

x = 99,99 – 5,678

x = 94.312

c) x x 5 = 105,85

x = 105,85 : 5

x = 21.17

d) x : 8,9 = 9,8

x = 9,8 x 8,9

x = 87.22

e) 76,32 : x = 36

x = 76,32 :  36

x = 2.12

a,x=(1,345.13)-15,96=1,525

b, x=99,99-6,678=93,312

c,x=105,85:5=21,17

d, x=9,8x8,9=87,22

e, x=76,32:36=2,12

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

\(2\left(x+3\right)+x\left(3+x\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

<=> (x+3)(x+2)=0

TH1 x+3=0 <=> x=-3

TH2 x+2=0 <=> x=-2

Vậy.... 

a: ta có: \(\left(x+3\right)\left(x-1\right)-x\left(x-5\right)=11\)

\(\Leftrightarrow x^2+2x-3-x^2+5x=11\)

\(\Leftrightarrow x=2\)

b: Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+2\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x^3-3x^2-2x+3x^2=0\)

\(\Leftrightarrow2x=64\)

hay x=32