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

Lời giải:
$a(x+2)^2+b(x+3)^3=cx+5$

$\Leftrightarrow bx^3+x^2(a+9b)+x(4a+27b)+(4a+27b)=cx+5$

Để điều này xảy ra với mọi $x\in\mathbb{R}$ thì:

\(\left\{\begin{matrix} b=0\\ a+9b=0\\ 4a+27b=c\\ 4a+27b=5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} b=0\\ a=0\\ c=0\\ 4a+27b=5\end{matrix}\right. \) (vô lý)

Do đó không tồn tại $a,b,c$ thỏa đề.

a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)

=>2x=-3/2

hay x=-3/4

b: 2x+3=5

=>2x=2

hay x=1

c: =>3(x-2)=4(5+x)

=>4x+20=3x-6

=>x=-26

a) => (7/2 + 2x) . 8/3 = 16/3
=> 7/2 + 2x = 16/3 : 8/3
=> 7/2 + 2x = 2
=> 2x = 2 - 7/2
=> 2x = -1.5
=> x = -1.5 : 2
=> x = -0.1

Bài 13: 

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

hay x=0

a) \(x+1^3=2^5-\left(-1^3\right)\)

\(\Rightarrow x+1=33\)

=> x = 32

b) \(3^7-x=1^4-\left(-3^5\right)\)

\(\Rightarrow2187-x=1+243=244\)

=> x = 1943

a) \(\Leftrightarrow x+1=32+1\)

\(\Leftrightarrow x=32\)

Vậy x = 32

b) \(\Leftrightarrow2187-x=1+243\)

\(\Leftrightarrow2187-x=244\)

\(\Leftrightarrow x=1943\)

Vậy x = 1943

a) \(x+1^3=2^5-\left(-1^3\right)\) 

      \(x+1=32-\left(-1\right)\) 

      \(x+1=33\) 

            \(x=33-1\) 

            \(x=32\) 

b) \(3^7-x=1^4-\left(-3^5\right)\) 

\(2187-x=1-\left(-243\right)\) 

\(2187-x=244\) 

            \(x=2187-244\) 

            \(x=1943\)

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

`a)sqrt{9x^2}=6`

`<=>|3x|=6`

`<=>|x|=2`

`<=>x=+-2`

`b)sqrt{(x-2)^2}=5`

`<=>|x-2|=5`

`**x-2=5`

`<=>x=7`

`**x-2=-5`

`<=>x=-3`

`c)sqrt{x^2-6x+9}=3`

`<=>\sqrt{(x-3)^2}=3`

`<=>|x-3|=3`

`**x-3=3`

`<=>x=6`

`**x-3=-3`

`<=>x=0`

`d)sqrt{x^2+4x+4}-2x=3`

`<=>sqrt{(x+2)^2}=3+2x`

`<=>|x+2|=2x+3(x>=-3/2)`

`**x+2=2x+3`

`<=>x=-1(tm)`

`**x+2=-2x-3`

`<=>3x=-5`

`<=>x=-5/3(l)`

Sử dụng công thức:`sqrtA^2=|A|`

ĐKXĐ : \(x\in R\)

a, \(\sqrt{9x^2}=\left|3x\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)

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

Vậy ..

b, \(\sqrt{\left(x-2\right)^2}=\left|x-2\right|=5\)

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

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

Vậy ...

c, \(\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|=3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

Vậy ..

d, \(\sqrt{x^2+4x+4}-2x=\sqrt{\left(x+2\right)^2}-2x=\left|x+2\right|-2x=3\)

\(\Leftrightarrow\left|x+2\right|=2x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=2x+3\\x+2=-2x-3\end{matrix}\right.\\2x+3\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{2}\\\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-\dfrac{5}{3}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy ..