a)
A = \(\left(\dfrac{x}{x-2}-\dfrac{x^3}{x^3-8}\right):\dfrac{x^2+2x}{x^2+2x+4}\)
A=\(\left[\dfrac{x}{x-2}-\dfrac{x^3}{\left(x-2\right)\left(x^2+2x+4\right)}\right].\dfrac{x^2+2x+4}{x^2+2x}\)
A = \(\left[\dfrac{x\left(x^2+2x+4\right)-x^3}{\left(x-2\right)\left(x^2+2x+4\right)}\right].\dfrac{x^2+2x+4}{x^2+2x}\)
A = \(\dfrac{x^3+2x^2+4x-x^3}{\left(x-2\right)\left(x^2+2x+4\right)}.\dfrac{x^2+2x+4}{x^2+2x}\)
A = \(\dfrac{2x^2+4x}{\left(x-2\right)\left(x^2+2x\right)}\)
A = \(\dfrac{2\left(x^2+2x\right)}{\left(x-2\right)\left(x^2+2x\right)}=\dfrac{2}{x-2}\)
b) ĐKXĐ: x \(\ne\) 2
Để A > 3
thì \(\dfrac{2}{x-2}>3\)
\(2>3\left(x-2\right)\)
\(2>3x-6\)
\(-3x>-8\)
\(x< \dfrac{8}{3}\) ( thoả mãn ĐKXĐ )
Vậy để A >3 thì \(x< \dfrac{8}{3}\)
a)
\(A=\left[\dfrac{x}{x-2}-\dfrac{x^3}{x^3-8}\right]:\dfrac{x^2+2x}{x^2+2x+4}\\ A=\left[\dfrac{x}{x-2}-\dfrac{x^3}{\left(x-2\right)\left(x^2+2x+4\right)}\right]:\dfrac{x\left(x+2\right)}{x^2+2x+4}\\ A=\dfrac{x\left(x^2+2x+4\right)-x^3}{\left(x-2\right)\left(x^2+2x+4\right)}:\dfrac{x\left(x+2\right)}{x^2+2x+4}\\ A=\dfrac{2x\left(x+2\right)\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)x\left(x+2\right)}\\ A=\dfrac{2}{x-2}\)
b)ĐKXĐ: \(x\ne2\)
\(A>3\Rightarrow\dfrac{2}{x-2}>3\\ 2>3x-6\\ 8>3x\\ \dfrac{8}{3}>x\)
vậy khi \(\left\{{}\begin{matrix}\dfrac{8}{3}>x\\x\ne2\end{matrix}\right.\) thì A>3
