Bài 8: Phép chia các phân thức đại số

Để x^3 +x^2 +a-x chia hết cho (x+1)^2 thì a+1 = 0 hay a=-1

Vậy....(cái gì chia hết cho cái gì đó <=> a=-1)

Đặt tính ra được: a= -1

\(\left(x-y-z\right)^5:\left(x-y-z\right)^3=\left(x-y-z\right)^{5-3}=\left(x-y-z\right)^2\)

x khác 0 và x khác cộng trừ 1

a: Để \(5x^3y^7⋮x^ny^4\) thì 3-n>=0

hay n<=3

b: Để \(8x^4y^3⋮x^{2n}y^n\) thì \(\left\{{}\begin{matrix}4-2n>=0\\3-n>=0\end{matrix}\right.\Leftrightarrow n< =2\)

c: Để \(15x^ny^5⋮3x^2y^n\) thì \(\left\{{}\begin{matrix}n-2>=0\\5-n>=0\end{matrix}\right.\Leftrightarrow2< =x< =5\)

\(=\left(\dfrac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right):\dfrac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)

\(=\dfrac{4a+2b-6b-8a+4b}{8a^2}\)

\(=\dfrac{-4a}{8a^2}=\dfrac{-1}{2a}\)

ta có:

(\(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\)):\(\dfrac{2x-6}{x^2+6x}\)+\(\dfrac{x}{6-x}\)

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

= (\(\dfrac{x^2}{x\left(x-6\right)\left(x+6\right)}-\dfrac{\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\)).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{x^2-x^2+12x-36}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12x-36}{x\left(x-6\right)\left(x+6\right)}\). \(\dfrac{x^2+6x}{2x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\).\(\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)+\(\dfrac{x}{6-x}\)

= \(\dfrac{6}{x-6}\)- \(\dfrac{x}{x-6}\)

= \(\dfrac{6-x}{x-6}\)

= \(\dfrac{-\left(x-6\right)}{x-6}\)

= -1

Gọi thương khi chia \(f\left(x\right)\) cho \(x^2-4\) là \(Q\left(x\right)\), ta có;

\(x^4+ax+b=\left(x+2\right)\left(x-2\right).Q\left(x\right)\)

Vì đẳng thức đúng với mọi \(x\) nên lần lượt cho \(x=-2,x=2\), ta được:

\(\left\{{}\begin{matrix}16-2a+b=0\\16+2a+b=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-2a+b=-16\\2a+b=-16\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=0\\b=-16\end{matrix}\right.\)

Vậy với \(a=0;b=-16\) thì \(f\left(x\right)⋮x^2-4\)

x=-1 ,-5

a) (x³ - 11x +5 - 3x²) : (x - 5)

= (x³ - 3x² - 11x + 5) : (x - 5)

= (x³ - 5x² + 2x² - 10x - x + 5) : (x - 5)

= [(x³ - 5x²) + (2x² - 10x) - (x - 5)] : (x - 5)

= [x²(x - 5) + 2x(x - 5) - (x - 5)] : (x - 5)

= (x² + 2x - 1)(x - 5) : (x - 5)

= x² + 2x - 1

b) (4x⁴ - 5x² - 3 + 3x³ + 9x): (x² - 3)

con b bn xem có chép nhầm đề bài ko nhé

a, (x³-11x+5-3x²) : (x-5)= (x³-3x²-11x+5) : (x-5)

Tính :

b, (4x⁴-5x²-3-3x³+9x) : (x²-3)= (4x⁴-3x³-5x²+9x-3) : (x²-3)

Tính: