Bài 9: Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

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

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

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

a, DKXD: \(\left\{{}\begin{matrix}x+2\ne0\\x-2\ne0\\x^2+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)

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

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

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

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

Lời giải:

$f(x)=x^4+6x^3+7x^2-6x+a=x^2(x^2+3x-1)+3x(x^2+3x-1)-(x^2+3x-1)+a-1$
$=(x^2+3x-1)(x^2+3x-1)+(a-1)$
Vậy $f(x)$ chia $x^2+3x-1$ dư $a-1$

Để đây là phép chia hết thì $a-1=0\Leftrightarrow a=1$

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

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

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

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

\(=-x^2+3\)

a) \(ĐKXĐ:x\ne\pm1\)

b) \(M=\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right).\dfrac{x^2-1}{15}\)

\(=\left[\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\dfrac{6}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right].\dfrac{\left(x-1\right)\left(x+1\right)}{15}\)

\(=\left[\dfrac{x^2+2x+1+6-\left(x^2+2x-3\right)}{2\left(x-1\right)\left(x+1\right)}\right].\dfrac{\left(x-1\right)\left(x+1\right)}{15}\)

\(=\left(\dfrac{4}{2\left(x-1\right)\left(x+1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{15}=\dfrac{2}{15}\)

Vậy...

b: \(=\dfrac{3x^3+6x^2-9x-6x^2-12x+18+20x-16}{x^2+2x-3}\)

\(=3x-6+\dfrac{20x-16}{x^2+2x-3}\)

c: \(=\dfrac{3x^3-6x^2+4x^2-8x+12x-24+28}{x-2}\)

\(=3x^2+4x+12+\dfrac{28}{x-2}\)

d: \(=\dfrac{x^3-2x^2+3x^2-12}{x-2}=x^2+3x+6\)

a, ĐKXĐ:\(2x^3-2x^2\ne0\Rightarrow2x^2\left(x-1\right)\ne0\Rightarrow\left\{{}\begin{matrix}2x^2\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne1\end{matrix}\right.\)

b, \(A=\dfrac{5x^2-5x}{2x^3-2x^2}\)

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

\(\Rightarrow A=\dfrac{5}{2x}\)

Để A=1\(\Rightarrow\dfrac{5}{2x}=1\)

\(\Rightarrow2x=5\\ \Rightarrow x=\dfrac{5}{2}\)

a, đk \(2x^2\left(x-1\right)\ne0\Leftrightarrow x\ne0;x\ne1\)

b, \(A=\dfrac{5x\left(x-1\right)}{2x^2\left(x-1\right)}=\dfrac{5}{2x}=1\Rightarrow5=2x\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

a) Để A xác định thì 

\(2x^3-2x^2\ne0\Leftrightarrow2x^2\left(x-1\right)\ne0\Leftrightarrow x\ne0;1\)

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

Tại x=1 , giá trị của A là :\(\dfrac{5}{2\cdot1}=\dfrac{5}{2}\)