Phân thức đại số

3x² - 6x = 3x.(x - 2)

x² - 2x + 1 - y²= (x - 1)² - y² = (x + y -1).(x - y + 1)

9x³ - 9x²y - 4x +4y =(9x³ - 9x²y) - (4x - 4y) =9x.(x - y) - 4.(x - y) =(x- y).(9x - 4)

x³ - 2x² - 8x = x.(x² - 2x - 8) = x.(x² - 4x + 2x -8) = x.[x.(x - 4) + 2.(x - 4)] =

x.(x - 4).(x + 2)

1. x³ - 2x² - 8x

= x (x² - 2x - 8)

= x (x² - 4x + 2x - 8)

= x [x(x - 4) + 2 (x - 4)]

= x (x - 4) (x + 2)

2. 3x² - 6x

= 3x (x - 2)

3. x² - 2x + 1 - y²

= (x + 1)² - y²

= (x + 1 + y) (x + 1 - y)

4. 9x³ - 9x² - 4x + 4y

= 9x² (x - y) - 4 (x - y)

= (x - y) (9x² - 4)

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

mình cần gấp giúp mình với

a:TXĐ D=R\{2}

b: \(P=\dfrac{x^2}{x^3-8}+\dfrac{x}{x^2+2x+4}+\dfrac{1}{x-2}\)

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

\(=\dfrac{3x^2+4}{x^3-8}\)

\(a,9a^2b+12ab^3-18ab\)

\(=3ab\left(3a+4b^2-6\right)\)

\(b,x^2+2xy-25+y^2\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

\(=\left(x+y-5\right)\left(x+y+5\right)\)

a. 9a²b + 12ab³ - 18ab

= 3ab(3a + 4b² - 6)

b. x² + 2xy - 25 + y²

= (x² + 2xy + y²) - 25

= (x + y)² - 5²

= (x + y + 5)(x + y - 5)

\(\Rightarrow x^2-x-5x+5=0\)

\(\Rightarrow\left(x^2-x\right)-\left(5x-5\right)=0\)

\(\Rightarrow x.\left(x-1\right)-5.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right).\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow x^2-x-5x+5=0\\ \Leftrightarrow x\left(x-1\right)-5\left(x-1\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Pt có dạng a + b + c = 1 + (-6) + 5 = 0

Nên ta có 

x1 = 1 

x2 = c/a = 5

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

a) \(x^2+2xy+y^2-4=\left(x+y\right)^2-2^2\)

\(=\left(x+y-2\right)\left(x+y+2\right)\)

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

\(=\left(x+y\right)\left(x-y+1\right)\)

c) \(y^2+x^2+2xy-16=x^2+2xy+y^2-16\)

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

\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)

\(\left\{{}\begin{matrix}\dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\\dfrac{x-3}{x}\\\dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\end{matrix}\right.\)

Vậy \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

\(ĐK:x\ne0;x\ne\pm1\\ \dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\ \dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\)

Do đó 3 phân thức trên bằng nhau

Giúp mik vs ạ