Câu 1:
a: Đặt biểu thức cần tìm là A
Theo đề, ta có: \(\dfrac{x^3-2^3}{2x\left(x-2\right)}=\dfrac{2x^2+4x+8}{A}\)
=>\(\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{2x\left(x-2\right)}=\dfrac{2\left(x^2+2x+4\right)}{A}\)
=>\(\dfrac{x^2+2x+4}{2x}=\dfrac{2\left(x^2+2x+4\right)}{A}\)
=>\(A=\dfrac{2x\cdot2\left(x^2+2x+4\right)}{x^2+2x+4}=4x\)
b: Đặt biểu thức cần tìm là A
Theo đề, ta có: \(\dfrac{A}{x^2-4x+4}=\dfrac{x^2+2x}{x^2-4}\)
=>\(\dfrac{A}{\left(x-2\right)^2}=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>\(\dfrac{A}{\left(x-2\right)^2}=\dfrac{x}{x-2}\)
=>\(A=\dfrac{x\left(x-2\right)^2}{x-2}=x\left(x-2\right)=x^2-2x\)
c: Gọi đa thức cần tìm là B
Theo đề, ta có: \(\dfrac{B}{3x-2}=\dfrac{2x^2-3x}{3x^2-2x}\)
=>\(\dfrac{B}{3x-2}=\dfrac{x\left(2x-3\right)}{x\left(3x-2\right)}=\dfrac{2x-3}{3x-2}\)
=>B=2x-3
d: Gọi đa thức cần tìm là A
Theo đề, ta có: \(\dfrac{2\left(x+y\right)^2}{A}=\dfrac{4x+4y}{3}\)
=>\(A=\dfrac{3\cdot2\left(x+y\right)^2}{4x+4y}=\dfrac{6\left(x+y\right)^2}{4\left(x+y\right)}=\dfrac{3}{2}\left(x+y\right)\)
Câu 2:
a: \(\dfrac{8x^3}{-2x^2}=\dfrac{8x^3:\left(-2x\right)}{-2x^2:\left(-2x\right)}=\dfrac{-4x^2}{x}=\dfrac{-4x^2\cdot x^2}{x\cdot x^2}=\dfrac{-4x^4}{x^3}\)
b: \(\dfrac{x-2}{2x-4}=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(2x-4\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-2x-4}{2x^2+4x-4x-8}\)
\(=\dfrac{x^2-4}{2x^2-8}\)
c: \(\dfrac{-2x^4y^3}{6xy}=\dfrac{-x^5y^4}{3x^2y^2}\)
=>\(-2x^4y^3\cdot3x^2y^2=-x^5y^4\cdot6xy\)
=>\(-6x^6y^5=-6x^6y^5\)(Đúng)
d: \(\dfrac{x^2+xy+y^2}{2xy}=\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{2xy\left(x-y\right)}\)
\(=\dfrac{x^3-y^3}{2x^2y-2xy^2}\)
