Bài 8: Phép chia các phân thức đại số

PTĐTTNT?

\(x^2-2xy+y^2+2x-2y\)

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

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

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

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

x=-1 ,-5

\(A=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{5\left(x-1\right)}{2x}=\dfrac{10}{x+1}\)

Để A nguyên thì \(x-3\in\left\{-1;1;2;-2\right\}\)

hay \(x\in\left\{2;4;5;1\right\}\)

Đợi hết 5 phút đi r t giải cho =))

x khác 0 và x khác cộng trừ 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 \(f\left(x\right)⋮x^2-4\)

b: \(x^2\left(x-2x^3\right)=x^3-2x^5\)

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

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

c: \(\left(x-2\right)\left(x^2+3x-4\right)\)

\(=x^3+3x^2-4x-2x^2-6x+8\)

\(=x^3+x^2-10x+8\)

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

\(A=\left(\dfrac{x}{y^2-xy}+\dfrac{y}{x^2-xy}\right):\left(\dfrac{x^2+y^2}{xy^2+x^2y}\right)\)

\(A=\left(\dfrac{x}{y\left(y-x\right)}+\dfrac{y}{x\left(x-y\right)}\right):\left(\dfrac{x^2+y^2}{xy\left(x+y\right)}\right)\\ \)

\(x,y\ne0;\left|y\right|\ne\left|x\right|\)

\(\)\(A=\left(\dfrac{x}{y\left(y-x\right)}+\dfrac{y}{x\left(x-y\right)}\right).\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

\(A=\left(\dfrac{x}{y\left(y-x\right)}.\dfrac{xy}{x^2+y^2}+\dfrac{y}{x\left(x-y\right)}.\dfrac{xy}{x^2+y^2}\right)\left(x+y\right)\)

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

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

\(\)

\(\left(\dfrac{x}{-y\left(x-y\right)}+\dfrac{y}{x\left(x-y\right)}\right).\dfrac{xy\left(x+y\right)}{x^2+y^2}\)

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

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

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

giúp mik với ạ , mai nộp rồi

c: \(\dfrac{27x^5y^3}{9x^3y^2}=3x^2y\)

\(=3\left(-3\right)^2\cdot2=3\cdot9\cdot2=54\)