\(A\left(x\right)-B\left(x\right)=\left(2x^3-3x^2+3x+8\right)-\left(3x^3+2x^3-5x+1\right)\\ =2x^3-3x^2+3x+8-3x^3-2x^3+5x-1\\ =\left(2x^3-3x^3+2x^3\right)+\left(-3x^2\right)+\left(3x+5x\right)+\left(8-1\right)\\ =x^3-3x^2+8x+7\)

a: 3x-7=8

=>3x=15

hay x=2

b: =>2x+5=3x-15

=>-x=-20

hay x=20

c: =>3(3x-4)=2(4x+1)

=>9x-12=8x+2

=>x=14

a) 3x-7=8
     3x=15
       x=3
Vậy..
b) 5+2x=3(x-5)

     2x-3x=-5-15
     -x=-20
      x=20
Vậy...

c) 3x -4/2=4x+1/3

    3x-4x=2+1/3
    -x=7/3
     x=-7/3
Vậy...

a

Chọn A

2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

a) \(\left(2x-1\right)^{10}=49^5\Leftrightarrow\left(2x-1\right)^{10}=7^{10}\Leftrightarrow\orbr{\begin{cases}2x-1=7\Rightarrow x=4\\2x-1=-7\Rightarrow x=-3\end{cases}}\)

PT có 2 nghiệm: x = -3 và x = 4.

b) \(3^x+2+3x=810\Leftrightarrow3^x+3x=808\)(2)

x = 0 không phải là nghiệm của (2)

VT(2) chia hết cho 3 với mọi x khác 0; => PT vô nghiệm

a: Ta có: \(27x^3-54x^2+36x=8\)

\(\Leftrightarrow27x^3-54x^2+36x-8=0\)

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

\(\Leftrightarrow3x-2=0\)

hay \(x=\dfrac{2}{3}\)

b: Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+5\right)=x^2+3x\)

\(\Leftrightarrow\left(x+3\right)\cdot\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x=-3\)

a.

\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

b.

\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)

c.

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

d.

\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

e.

\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)

f.

\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(3^{x+2}-3^x=72\)

\(\Rightarrow3^x\cdot\left(3^2-1\right)=72\)

\(\Rightarrow3^x\cdot\left(9-1\right)=72\)

\(\Rightarrow3^x\cdot8=72\)

\(\Rightarrow3^x=\dfrac{72}{8}\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

_____

Xem lại đề 

a) Ta có : x=0 không là nghiệm của phương trình.          Chia cả hai vế của phương trình cho \(^{x^2}\) ta có:

    \(x^2-2x-1-\frac{2}{x}+\frac{1}{x^2}=0\) \(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)-1=0\)  (1)

Đặt \(x+\frac{1}{x}=t\)  \(\left(t>2\right)\) hoăc \(\left(t

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x^2-2.x.3+3^2\right)=2x\left(x-3\right)^2\)