Question

[nguyen ngoc son]

cho biểu thức P=\(\left(\dfrac{a}{3a^2-6a}+\dfrac{2a-3}{2a^2-a^3}\right).\dfrac{6a}{a^2-6a+9}\)

a.rút gọn P

b.tìm giá trị của A để P>0

Answer 1

[Akai Haruma]

Lời giải:

a) ĐKXĐ: $a\neq 0; a\neq 3; a\neq 2$

\(P=\left[\frac{a}{3a(a-2)}-\frac{2a-3}{a^2(a-2)}\right].\frac{6a}{(a-3)^2}=\left[\frac{a^2}{3a^2(a-2)}-\frac{6a-9}{3a^2(a-2)}\right].\frac{6a}{(a-3)^2}=\frac{a^2-6a+9}{3a^2(a-2)}.\frac{6a}{(a-3)^2}=\frac{(a-3)^2}{3a^2(a-2)}.\frac{6a}{(a-3)^2}=\frac{2}{a(a-2)}\)

b) 

Để $P>0\Leftrightarrow \frac{2}{a(a-2)}>0\Leftrightarrow a(a-2)>0$

$\Leftrightarrow a>2$ hoặc $a< 0$

Kết hợp với ĐKXĐ suy ra $(a>2; a\neq 3)$ hoặc $a< 0$

Answer 2

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: \(a\notin\left\{0;2\right\}\)

a) Ta có: \(P=\left(\dfrac{a}{3a^2-6a}+\dfrac{2a-3}{2a^2-a^3}\right)\cdot\dfrac{6a}{a^2-6a+9}\)

\(=\left(\dfrac{a}{3a\left(a-2\right)}+\dfrac{2a-3}{a^2\left(2-a\right)}\right)\cdot\dfrac{6a}{a^2-6a+9}\)

\(=\left(\dfrac{a^2}{3a^2\cdot\left(a-2\right)}-\dfrac{3\left(2a-3\right)}{3a^2\cdot\left(a-2\right)}\right)\cdot\dfrac{6a}{\left(a-3\right)^2}\)

\(=\dfrac{a^2-6a+9}{3a^2\cdot\left(a-2\right)}\cdot\dfrac{6a}{\left(a-3\right)^2}\)

\(=\dfrac{\left(a-3\right)^2}{3a^2\left(a-2\right)}\cdot\dfrac{6a}{\left(a-3\right)^2}\)

\(=\dfrac{2}{a\left(a-2\right)}\)

b) Để P>0 thì \(\dfrac{2}{a\left(a-2\right)}>0\)

mà 2>0

nên \(a\left(a-2\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\a-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\a-2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\a>2\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\a< 2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)

Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)

Vậy: Để P>0 thì \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)