hello

a, \(P=\left(\frac{x^2+9}{x^2+5x}+\frac{x-1}{x}-\frac{x}{x+5}\right)\left(1+\frac{2}{x}\right)\)đk : x khác  0 ; -5 

\(=\left(\frac{x^2+9+x^2+4x-5-x^2}{x\left(x+5\right)}\right)\left(\frac{x+2}{x}\right)\)

\(=\frac{x^2+4x+4}{x\left(x+5\right)}\left(\frac{x+2}{x}\right)=\frac{\left(x+2\right)^3}{x^2\left(x+5\right)}\)

b, Ta có \(\left(x+2\right)\left(3x-2\right)=0\Leftrightarrow x=-2;x=\frac{2}{3}\)

Với x = -2 => P = 0 

Với x = 2/3 => \(P=\frac{\left(\frac{2}{3}+2\right)^3}{\frac{4}{9}\left(\frac{2}{3}+5\right)}=\frac{128}{17}\)

-mình nghĩ bạn nên đặt dấu chia giữa 2 đa thức kia thì kq sẽ đẹp hơn 

3x(x – 2) – 5x(1 – x) – 8( x 2  – 3)

= 3x.x + 3x .( -2) – [5x.1 + 5x. (- x)] – [8 x 2  + 8.(- 3)]

= (3 x 2  – 6x) – (5x – 5 x 2 ) – (8 x 2  – 24)

= 3 x 2  – 6x – 5x + 5 x 2  – 8 x 2 + 24

= ( 3 x 2  +5 x 2  – 8 x 2 )- ( 6x + 5x) + 24

= - 11x + 24

\(a,x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

\(=-3x^3-3x\)

\(b,3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=-11x+24\)

=.= hok tốt!!

a)

\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2=x\left(2x^2-3-5x^2-x+x\right)\)

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

b)

Nhân ra:

\(3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)

Chúc bạn học tốt!

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