Bài 1: Phân thức đại số.

Bài 1:

\(a,=\dfrac{4x-8}{x-2}=\dfrac{4\left(x-2\right)}{x-2}=4\\ c,=\dfrac{3x+1-2x}{x\left(x+1\right)}=\dfrac{x+1}{x\left(x+1\right)}=\dfrac{1}{x}\\ e,=\dfrac{2x-3+2x^2+2x-3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+x}{\left(x-1\right)\left(x+1\right)}\\ b,=\dfrac{x-13+x+3}{x-5}=\dfrac{2\left(x-5\right)}{x-5}=2\\ d,=\dfrac{x+4+x-4-3x}{x\left(x-4\right)}=\dfrac{-x}{x\left(x-4\right)}=\dfrac{1}{4-x}\\ f,=\dfrac{x+8-16+2x-16}{\left(x-8\right)\left(x+8\right)}=\dfrac{3x-24}{\left(x-8\right)\left(x+8\right)}=\dfrac{3\left(x-8\right)}{\left(x-8\right)\left(x+8\right)}=\dfrac{3}{x+8}\)

Bài 1: 

b: \(=\dfrac{x-13+x+3}{x-5}=2\)

f: \(=\dfrac{x+8-16+2x-16}{\left(x-8\right)\left(x+8\right)}=\dfrac{3x-24}{\left(x-8\right)\left(x+8\right)}=\dfrac{3}{x+8}\)

1, x^3-2x^2+x

=x^3-x^2-x^2+x

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

= x^2(x-1)-x(x-1)

=(x-1)(x^2-x)

=x(x-1)(x-1)

2, x^2-2x-15

=x^2-2x-9-6

= x^2-9-2x-6

=(x^2-9)-(2x+6)

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

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

3 , \(^{3x^3y^2-6x^2y^3+9x^2y^2}\)

= \(^{3x^2y^2\left(x-2y+3\right)}\)

4,  \(^{5x^2y^3-25x^3y^4+10x^3y^3}\)

=\(^{5x^2y^2\left(y-5xy^2+2xy\right)}\)

5, \(^{12x^2y-18xy^2-30y^2}\)

=\(^{3y\left(4x^2-6xy-10y\right)}\)

Lời giải:
1. $x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$
2. $x^2-2x-15=(x^2+3x)-(5x+15)=x(x+3)-5(x+3)=(x+3)(x-5)$

3. $3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2(x-2y+3)$

4. $5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3(1-5xy+2x)$
5. $12x^2y-18xy^2-30y^2=6y(2x^2-3xy-5y)$

a)\(x^2-3=\left(x-3\right)x\)

b)\(x^2-36=\left(x-6\right)^2=\left(x-6\right)\left(x+6\right)\)

c)\(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)^2\)

d)\(x^2-4x+4=x^2-2x2+4=\left(x-2\right)^2=\left(x-2\right)\left(x+2\right)\)

e)\(5x+x^2=x\left(5+x\right)\)

f)\(4x^2-9=\left(2x\right)^2-9=\left(2x-3\right)\left(2x+3\right)\)

h)\(4x-x^2=x\left(4-x\right)\)

i\(x^2+6x+9=x^2+2x3+9=\left(x+3\right)^2=\left(x+3\right)\left(x-3\right)\)

\(c,ĐK:x\ne\dfrac{5}{4}\\ \dfrac{2x+3}{4x-5}=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\\ d,ĐK:x\ne1\\ \dfrac{x^2-1}{x^2-2x+1}=0\\ \Leftrightarrow x^2-1=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=-1\)

c, ĐK:x khác  5/4

Ta có 2x+3/4x-5=0=> 2x +3 =0=> x=-3/5

d, ta có x^2-1/x^2-2x+1=x+1/x-1 (ĐK : x khác 1)

Để x^2-1/x^2-2x+1=0 =>x+1/x-1=0

=>x=-1

\(\dfrac{10pq\left(2q-1\right)^3}{15p^3-30p^4}=\dfrac{-10pq\left(1-2p\right)^3}{15p^3\left(1-2p\right)}=\dfrac{-2q\left(1-2p\right)^2}{3p^2}\)