Bài 8: Phép chia các phân thức đại số

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

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

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

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

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

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

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

\(\)

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

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

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

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

a: \(=\dfrac{x-2}{x-5}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)^2}=\dfrac{x+5}{x-2}\)

b: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)

c: \(=\dfrac{a+6-3}{3\left(a+3\right)}\cdot\dfrac{27a}{a+2}=\dfrac{a+3}{3\left(a+3\right)}\cdot\dfrac{27a}{a+2}\)

\(=\dfrac{9a}{a+2}\)

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

\(=\dfrac{-x-2}{5x}\)

b: Để P=x² thì \(5x^3+x+2=0\)

hay \(x\in\left\{-0.65\right\}\)

c: Để |P|<P thì \(x\in\varnothing\)

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

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

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

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

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

b) Bạn có thể viết kiểu latex được không ạ ?

Ta có: \(x^3-5x-1:x+2\)

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

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

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

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

Ta có: \(\left(\dfrac{3x}{1-2x}+\dfrac{2x}{2x+1}\right):\dfrac{2x^2+5x}{1-4x+4x^2}\)

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

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

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

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

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

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

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

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

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

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

= \(\dfrac{-3}{x-2}\)

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

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

a) Đkxđ:  \(x-2\ne0\Leftrightarrow x\ne2\)

               \(x+2\ne0\Leftrightarrow x\ne-2\)

               \(2-x\ne0\Leftrightarrow x\ne2\)

               \(x-3\ne0\Leftrightarrow x\ne3\)

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

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

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

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

Chắc là mình làm sai rồi. Mình không làm nữa :v. Sorry bạn nha :<< 

Bài 8: 

a) ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)

b) Ta có: \(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)

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

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

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

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

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

c) Ta có: |x-5|=2

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Thay x=7 vào biểu thức \(P=\dfrac{x\left(-4x^2+8x\right)}{\left(x+2\right)\left(x-3\right)}\), ta được:

\(P=\dfrac{7\cdot\left(-4\cdot7^2+8\cdot7\right)}{\left(7+2\right)\left(7-3\right)}=\dfrac{7\cdot\left(-4\cdot28+56\right)}{9\cdot4}\)

\(\Leftrightarrow P=\dfrac{7\cdot\left(-112+56\right)}{36}=\dfrac{7\cdot\left(-56\right)}{36}=\dfrac{-98}{9}\)

Vậy: Với |x-5|=2 thì \(P=-\dfrac{98}{9}\)