Mn giúp mik vs ạ

1,

\(x^2-2ax+a^2=\left(x-a\right)^2\)

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

Vậy MSC: \(\left(x-a\right)^2x\)

2,

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

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

\(x^2+x+1\)

vậy MSC là: \(x\left(x-1\right)\left(x^2+x+1\right)\)

a)

\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)

\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\)

MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)

\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)

\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)

M.n jup mk câu b,c vs ạk

a) MTC: \(12x^3y^3\)

\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)

\(\dfrac{2}{3xy^3}=\dfrac{2\cdot4x^2}{3xy^3\cdot4x^2}=\dfrac{8x^2}{12x^3y^3}\)

b) MTC: \(x\left(x-3\right)^2\)

\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)

\(\dfrac{3}{x^2-3x}=\dfrac{3}{x\left(x-3\right)}=\dfrac{3\left(x-3\right)}{x\left(x-3\right)^2}=\dfrac{3x-9}{x\left(x-3\right)^2}\)

a) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}=\dfrac{ax\left(a^3-x^3\right)}{a^2+ax+x^2}=\dfrac{ax\left(a-x\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}=ax\left(a-x\right)\)

Thay a=3;x=1/2 vào biểu thức trên, ta có:

\(ax\left(a-x\right)=3.\dfrac{1}{2}\left(3-\dfrac{1}{2}\right)=\dfrac{3}{2}.\dfrac{5}{6}=\dfrac{5}{4}\)

a)\(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{-ax\left(a^3-x^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{-ax\left(a-x\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=-ax\left(a-x\right)\)

Thay \(a=3;x=\dfrac{1}{2}\) vào biểu thức ta có:

\(\left(-3.\dfrac{1}{2}\right).\left(3-\dfrac{1}{2}\right)\)

\(=-\dfrac{3}{2}.\dfrac{3}{2}\)

\(=-\dfrac{9}{4}\)

c)\(\dfrac{x^3+3x}{3x^3+x^5}=\dfrac{x\left(x^2+3\right)}{x^3\left(x^2+3\right)}=\dfrac{x}{x^3}=\dfrac{1}{x^2}\)

Thay \(x=\dfrac{-1}{2}\)vào biểu thức trên, ta có:

\(\dfrac{1}{x^2}=\dfrac{1}{\left(\dfrac{-1}{2}\right)^2}=\dfrac{1}{\dfrac{1}{4}}=4\)

b) x³³ hay x³ vậy bạn

a)

em 2k6, đọc phần lí thuyết r lm, nên có lỗi j sai mong mn thông cảm

bài 1,

a, \(3xy\left(4xy^2-5x^2y-4xy\right)\)

= \(3xy.4xy^2-3xy.5x^2y-3xy.4xy\)

=\(12x^2y^3-15x^3y^2-12x^2y^2\)

b, \(\left(2x-1\right)\left(4x^2+2x+1\right)\)

=\(\left(2x.4x^2+2x.2x+2x.1\right)-\left(1.4x^2+1.2x+1.1\right)\)

=\(8x^3+4x^2+2x-4x^2-2x-1\)

=\(8x^3+\left(4x^2-4x^2\right)+\left(2x-2x\right)-1\)

=\(8x^3-1\)