\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right).x^2.\left(1-2x\right)\)

\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)

\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)

\(=\left(x-2\right).1\)

\(=x-2\)

Ta có:

\(\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)x^2\left(1-2x\right)\)

\(=\left(x-2\right)\left(2x^3-x^2+1\right)+\left(x-2\right)\left(x^2-2x^3\right)\)

\(=\left(x-2\right)\left[\left(2x^3-x^2+1\right)+\left(x^2-2x^3\right)\right]\)

\(=\left(x-2\right)\left(2x^3-x^2+1+x^2-2x^3\right)\)

\(=\left(x-2\right).1\)

\(=x-2\)

(x – 2) . (2x³ – x² + 1) + (x – 2) x²(1 – 2x)

= (x – 2). [(2x³ – x² + 1) + x²(1 – 2x)]

= (x – 2). [2x³ – x² + 1 + x² . 1 + x² . (-2x)]

= (x – 2) . (2x³ – x² + 1 + x² – 2x³)

= (x – 2) .1

= x – 2

Bài 2: 

3x + 2(5 - x) = 0

<=> 3x + 10 - 2x = 0

<=> x + 10 = 0

<=> x = 0 - 10

<=> x = -10

=> x = -10

Bài 3: 

6(3q + 4q) - 8(5p - q) + (p - q)

= 6.3p + 6.4q - 8.5p - (-8).q + p - q

= 18p + 24q - 40p + 8q + p - q

= (18p - 40p + p) + (24q + 8q - q)

= -21p + 31q

a: Đặt A=3(5x-2)-|x-5|

TH1: x>=5

=>x-5>=0

A=3(5x-2)-|x-5|

=15x-6-(x-5)

=15x-6-x+5

=14x-1

TH2: x<5

=>x-5<0

A=3(5x-2)-|x-5|

=15x-6-(5-x)

=15x-6-5+x

=16x-11

b: Đặt B=|2x+3|+2x+7

TH1: x>=-3/2

=>2x+3>=0

B=|2x+3|+2x+7

=2x+3+2x+7

=4x+10

TH2: x<-3/2

=>2x+3<0

B=|2x+3|+2x+7

=-2x-3+2x+7

=4

c: đặt C=3x-1+|1-3x|

=3x-1+|3x-1|

TH1: x>=1/3

=>3x-1>=0

C=3x-1+|3x-1|

=3x-1+3x-1

=6x-2

TH2: x<1/3

=>3x-1<0

C=3x-1+|3x-1|

=3x-1+1-3x

=0

d: Đặt D=3(x-1)-2|x+3|

TH1: x>=-3

=>x+3>=0

D=3(x-1)-2|x+3|

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

=3x-3-2x-6

=x-9

TH2: x<-3

=>x+3<0

D=3(x-1)-2|x+3|

=3x-3-2(-x-3)

=3x-3+2x+6

=5x+3

\(2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\) 

\(x-2=0\Leftrightarrow x=2\)

Ta có bảng xét dấu: 

x                             \(\frac{-1}{2}\)                      2

2x+1            -              0             +                           +

x-2               -                             -                            +

*) Nếu \(x\le\frac{-1}{2}\)ta có phương trình

\(A=\left(-2x-1\right)-\left(-x+2\right)+1\)

\(A=-2x-1+x-2+1\)

\(A=-x-2\)

*) Nếu \(\frac{-1}{2}< x\le2\)ta có phương trình

\(A=\left(2x+1\right)-\left(-x+2\right)+1\)

\(A=2x+1+x+2+1\)

\(A=3x+4\)

*) Nếu \(x>2\)ta có phương trình

\(A=\left(2x+1\right)-\left(x-2\right)+1\)

\(A=2x+1-x+2+1\)

\(A=x+4\)

Vậy \(A=\hept{\begin{cases}-x-2\left(\frac{-1}{2}\le x\right)\\3x+4\left(\frac{-1}{2}< x\le2\right)\\x+4\left(x>2\right)\end{cases}}\)