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a)
2a^2+6a=2a(a+3) khác 0=> a khác 0 và a khác -3
a^2-9=(a-3)(a+3) khác 0=> a khác -3 và a khác 3
tỏng hợp a \(\ne\) {-3,0,3}
b)\(B=\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\frac{\left(a^2-9\right)-6a+18}{\left(a-3\right)\left(a+3\right)}=\frac{\left(a+3\right)^2.\left(a-3\right)^2}{2a.\left(a-3\right)\left(a+3\right)^2}=\frac{a-3}{2a}\)
c)B=0\(\frac{\left(a-3\right)}{2a}=0=>a=3\Rightarrow\left(loai\right)\) kết luận ko có giá trị nào a ;B =0
d)\(B=1\Rightarrow\left(a-3\right)=2a\Rightarrow a=-3\left(loai\right)\)không có giá trị nào của a cho B=1
\(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}\cdot\dfrac{1-6a-18}{a^2-9}\\ a,ĐK:a\ne0;a\ne\pm3\\ b,B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\dfrac{-17-6a}{\left(a-3\right)\left(a+3\right)}=\dfrac{-17-6a}{2a\left(a-3\right)}\\ c,B=0\Leftrightarrow-17-6a=0\Leftrightarrow a=-\dfrac{17}{6}\left(tm\right)\\ d,B=1\Leftrightarrow-17-6a=2a^2-6a\\ \Leftrightarrow2a^2=-17\Leftrightarrow a\in\varnothing\)
a) B xác định
\(\Leftrightarrow\begin{cases}2a^2+6a\ne0\\a^2-9\ne0\end{cases}\Leftrightarrow\begin{cases}2a\left(a+3\right)\ne0\\\left(a+3\right)\left(a-3\right)\ne0\end{cases}\Leftrightarrow\begin{cases}a\ne0\\a\ne-3\\a\ne3\end{cases}\)
Vậy để B xác định thì \(a\ne0\) và \(a\ne\pm3\)
b) \(B=\frac{\left(a+3\right)^2}{2a^2+6a}\cdot\left(1-\frac{6a-18}{a^2-9}\right)\)
\(=\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\cdot\frac{\left(a+3\right)\left(a-9\right)}{\left(a+3\right)\left(a-3\right)}\)
\(=\frac{a+3}{2a}\cdot\frac{a-9}{a+3}\)
\(=\frac{a-9}{2a}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}2a^2+6a\ne0\\a^2-9\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\left(a+3\right)\ne0\\\left(a-3\right)\left(a+3\right)\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a\ne0\\a-3\ne0\\a+3\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a\ne0\\a\ne3\\a\ne-3\end{matrix}\right.\)
b) \(B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(\dfrac{a^2-9}{a^2-9}-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a^2-9\right)-\left(6a-18\right)}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-9-6a+18}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{a^2-6a+9}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\dfrac{\left(a-3\right)^2}{a^2-9}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a-3\right)^2}{\left(a-3\right)\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a+3}{2a}.\dfrac{a-3}{a+3}\)
\(\Leftrightarrow B=\dfrac{\left(a+3\right)\left(a-3\right)}{2a\left(a+3\right)}\)
\(\Leftrightarrow B=\dfrac{a-3}{2a}\)
a) \(A=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{2}{2a-b}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{-2\left(b+2a\right)}{\left(b-2a\right)\left(b+2a\right)}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4\left(b-2a\right)}{\left(2a+b\right)\left(b-2a\right)}\right):\left(\frac{a\left(4a^2-b^2\right)}{4a^2-b^2}+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\frac{-2b-4a+6b-4b+8a}{\left(b-2a\right)\left(b+2a\right)}:\frac{4a^3-ab^2+4a^2+b^2}{4a^2-b^2}\)
\(=\frac{4a}{\left(b-2a\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=\frac{-4a}{\left(2a-b\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=.\frac{-4a}{4a^3-ab^2+4a^2+b^2}\)
b) ĐKXĐ: \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)
Ta thấy \(a=\frac{1}{3};b=2\)thỏa mãn điều kiện \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)nên thay vào A ta được:
bạn thay vào tự tính nhé mà cái phần rút gọn bạn vừa làm vừa check giùm bài mik nhé =)) sợ sai
a) \(ĐK:a\ne1;a\ne0\)
\(A=\left[\frac{\left(a-1\right)^2}{3a+\left(a-1\right)^2}-\frac{1-2a^2+4a}{a^3-1}+\frac{1}{a-1}\right]:\frac{a^3+4a}{4a^2}=\left[\frac{a^2-2a+1}{a^2+a+1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}\)\(=\left[\frac{a^3-3a^2+3a-1}{a^3-1}-\frac{1-2a^2+4a}{a^3-1}+\frac{a^2+a+1}{a^3-1}\right].\frac{4a^2}{a^3+4a}=\frac{a^3-1}{a^3-1}.\frac{4a}{a^2+4}=\frac{4a}{a^2+4}\)
b) Ta có: \(a^2+4\ge4a\)(*)
Thật vậy: (*)\(\Leftrightarrow\left(a-2\right)^2\ge0\)
Khi đó \(\frac{4a}{a^2+4}\le1\)
Vậy MaxA = 1 khi x = 2
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 \(\left[{}\begin{matrix}a>2\\a< 0\end{matrix}\right.\)
a) B = \(\frac{\left(a+3\right)^2}{2a^2+6a}\). \(\left(1-\frac{6a-18}{a^2-9}\right)\)
= \(\frac{\left(a+3\right)^2}{2a\left(a+3\right)}\). \(\left(1-\frac{6\left(a-3\right)}{\left(a-3\right)\left(a+3\right)}\right)\)
= \(\frac{a+3}{2a}\). \(\left(1-\frac{6}{a+3}\right)\)
= \(\frac{a+3}{2a}\). \(\frac{a+3-6}{a+3}\)
= \(\frac{a+3}{2a}\). \(\frac{a-3}{a+3}\)
= \(\frac{a-3}{2a}\)
b) B = \(\frac{a-3}{2a}\)= 1
\(\Leftrightarrow\)\(a-3=2a\)
\(\Leftrightarrow\)\(a=-3\)
Vậy khi B = 1 thì a = -3