Bài 3: Rút gọn phân thức

a, \(\dfrac{3\left|x-4\right|}{3x^2-3x-36}=\dfrac{3\left|x-4\right|}{3\left(x^2-x-12\right)}=\dfrac{\left|x-4\right|}{x^2-x-12}\)

b, \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}=\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)

\(=\dfrac{\left(-x-2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(x+8\right)}{x+2}=\dfrac{-x-8}{x+2}\)

c, \(\dfrac{x^2+5x+6}{x^2+4x+4}=1+\dfrac{x+2}{x^2+4x+4}=1+\dfrac{x+2}{\left(x+2\right)^2}\)

\(=1+\dfrac{1}{x+2}=\dfrac{x+3}{x+2}\)

a, tiếp:

+) Xét \(x\ge4\) có:

\(\dfrac{x-4}{x^2-x-12}=\dfrac{x-4}{x^2-4x+3x-12}=\dfrac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)

\(=\dfrac{x-4}{\left(x+3\right)\left(x-4\right)}=\dfrac{1}{x+3}\)

+) Xét x < 4 có:
\(\dfrac{4-x}{x^2-x-12}=\dfrac{4-x}{x^2-4x+3x-12}=\dfrac{4-x}{x\left(x-4\right)+3\left(x-4\right)}\)

\(=\dfrac{4-x}{\left(x+3\right)\left(x-4\right)}=\dfrac{-\left(x-4\right)}{\left(x+3\right)\left(x-4\right)}=\dfrac{-1}{x+3}\)

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

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

\(=\dfrac{\left(-2-x\right)\left(8+x\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(x+2\right)\left(8+x\right)}{\left(x+2\right)^2}\)

\(-\dfrac{8+x}{x+2}\)

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

\(\Leftrightarrow\dfrac{3^2-\left(x+5\right)^2}{\left(x+2\right)^2}\)

\(\Leftrightarrow\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)

\(\Leftrightarrow\dfrac{\left(-x-2\right)\left(x+8\right)}{\left(x+2\right)^2}\)

\(\Leftrightarrow\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}\)

\(\Leftrightarrow\dfrac{-\left(x+8\right)}{\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{-x-8}{x+2}.\)

\(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}\)

\(=\dfrac{x^2.\left(x^2-1\right)-4.\left(x^2-1\right)}{x^2.\left(x^2-1\right)-9.\left(x^2-1\right)}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)

\(=\dfrac{x^2-4}{x^2-9}\)

Chúc bạn học tốt!!!

\(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)

\(=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}\)

\(=\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)

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

Ta có: \(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)\(=\dfrac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\dfrac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}\)\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\).

a) \(\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

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

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

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

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

Ta thấy mẫu thức của phân thức vốn đã lớn hơn 0 với mọi x, vậy để p/t trên có giá trị bằng 0 thì tử thức phải bằng 0

\(\Rightarrow\left(x+1\right)^2\left(x^2-x+1\right)=0\)

\(\Rightarrow x=-1\)

Vậy x = -1

b) \(\dfrac{x^4-5x^2+4}{x^4-10x^2+9}\)

= \(\dfrac{x^4-x^3+x^3-x^2-4x^2+4}{x^4-x^3+x^3-x^2-9x^2+9}\)

= \(\dfrac{x^3\left(x-1\right)+x^2\left(x-1\right)-4\left(x-1\right)\left(x+1\right)}{x^3\left(x-1\right)+x^2\left(x-1\right)-9\left(x-1\right)\left(x+1\right)}\)

= \(\dfrac{\left(x-1\right)\left(x^3+x^2-4x-4\right)}{\left(x-1\right)\left(x^3+x^2-9x-9\right)}\)

= \(\dfrac{x^3+x^2-4x-4}{x^3+x^2-9x-9}\)

= \(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)-9\left(x+1\right)}\)

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

= \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\) ( ĐKXĐ : \(x\ne\pm3\) )

Để phân thức trên có giá trị bằng 0 thì tử thức phải bằng 0

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) ( thoả mãn điều kiện xác định )

Vậy x = 2 hoặc x = -2

a, \(\dfrac{4x^2-8xy}{10y-5x}=\dfrac{4x\left(x-2y\right)}{5\left(2y-x\right)}=\dfrac{-4x}{5}\)

b, \(\dfrac{\left(x-2\right)^2-1}{x^2-6x+9}=\dfrac{\left(x-2-1\right)\left(x-2+1\right)}{\left(x-3\right)^2}\)

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

c, \(\dfrac{x^2+8x+16}{x^2-16}=\dfrac{\left(x+4\right)^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\)

Vì \(x^2+1>0\) \(\Rightarrow\) Để A < 0 thì \(x+2< 0\)

=> \(x< -2\)

Vì tử là dương nên để A là số âm thì x + 2 phải là số âm.

=> x+2 < 0

=> x < -2

câu b là sao ko hiểu

a ) 4x^2 + 12x + 9

= ( 2x )^2 + 2.2x.3 + 3^2

= ( 2x + 3 ) ^2

Bài 2: 

a: \(P=\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

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

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

\(=\dfrac{1}{x+2}:\dfrac{4x+6-x+2-x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{1}{x+2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x+8}=\dfrac{x-2}{2x+8}\)

b: Để P=0 thì x-2=0

hay x=2(loại)

Để P=1 thì 2x+8=x-2

hay x=-10(nhận)

Để P>0 thì \(\dfrac{x-2}{2x+8}>0\)

=>x>2 hoặc x<-4