Bài 7: Phép nhân các phân thức đại số

10x² + 15x

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

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

Để C>0 thì x+1>0

hay x>-1

Câu 14: B

Câu 4: B

Câu 12:

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

a: \(=-\dfrac{4n^2}{12n}\cdot\dfrac{7m^3}{17m^4}=-\dfrac{n}{3}\cdot\dfrac{7}{17m}=\dfrac{-7n}{51m}\)

b: \(=\dfrac{3\left(b+2\right)}{\left(b-9\right)^3}\cdot\dfrac{2\left(b-9\right)}{\left(b+2\right)^2}\)

\(=\dfrac{6}{\left(b-9\right)^2\cdot\left(b+2\right)}\)

Lời giải:

$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$

$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$

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

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

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

\(=2-4=-2\)