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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 $a< 0$
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 \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
Ta có \(6a^2-15ab+5b^2=0\Leftrightarrow15ab=6a^2+5b^2\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{9a^2-b^2}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}=\dfrac{3a^2+6a^2+5b^2-6b^2}{9a^2-b^2}\)
\(Q=\dfrac{9a^2-b^2}{9a^2-b^2}=1\)
a)Thay x=2(TMDK) vào bt Q :
\(Q=\dfrac{2+1}{2^2-9}=-\dfrac{3}{5}\)
b) \(P=\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1-\left(x-1\right)\left(x+1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{2x^2-1-\left(x^2-1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{x^2+3x}{x\left(x+1\right)}=\dfrac{x\left(x+3\right)}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\)
c) \(M=P.Q=\dfrac{x+3}{x+1}.\dfrac{x+1}{x^2-9}\\ =\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x-3}\)
\(M=-\dfrac{1}{2}\\ =>\dfrac{1}{x-3}=-\dfrac{1}{2}\\ =>x-3=-2\\ =>x=1\left(TMDK\right)\)
ta có : \(x+3+\dfrac{4-3a^2}{a^2-9}=\dfrac{5}{2a^2+6a}\)
\(\Leftrightarrow x+3=\dfrac{5}{2a^2+6a}-\dfrac{4-3a^2}{a^2-9}\)
\(\Leftrightarrow x+3=\dfrac{5}{2a\left(a+3\right)}-\dfrac{4-3a^2}{\left(a+3\right)\left(a-3\right)}\) \(\Leftrightarrow x+3=\dfrac{5\left(a-3\right)-2a\left(4-3a^2\right)}{2a\left(a+3\right)\left(a-3\right)}\) \(\Leftrightarrow x+3=\dfrac{5a-15-8a+6a^3}{2a\left(a+3\right)\left(a-3\right)}=\dfrac{6a^3-3a-15}{2a\left(a+3\right)\left(a-3\right)}\)\(\Leftrightarrow x=\dfrac{6a^3-3a-15}{2a\left(a+3\right)\left(a-3\right)}-3=\dfrac{6a^3-3a-15-3.2a\left(a^2-9\right)}{2a\left(a+3\right)\left(a-3\right)}\)
\(\Leftrightarrow x=\dfrac{6a^3-3a-15-6a^3+54a}{2a\left(a+3\right)\left(a-3\right)}=\dfrac{51a-15}{2a\left(a^2-9\right)}\)
Bác làm nhanh ***** :((